
Class Q A 55 

Book : _2_3_fci_ 

CopigMI? H=d. 



COPYRIGHT DEPOSIT. 



PREVIOUSLY PUBLISHED 




ELEMENTS 2L 




PLANE GEOMETRY 




With Numerous Exercises 




By CHARLES N. SCHMALL 


and 


SAMUEL M. SHACK 




12 mo - - One- Half Leather - - $1.25, 


Net 



A FIRST COURSE 



IN 



ANALYTICAL GEOMETRY 

PLANE AND SOLID WITH NUMEROUS 
EXAMPLES 



BY 

CHARLES N. SCHMALL, B.A. 

INSTRUCTOR IN MATHEMATICS IN THE 
PUBLIC SCHOOLS OF NEW YORK 



SECOND EDITION, ENLARGED 




NEW YORK 

D. VAN NOSTRAND COMPANY 

EIGHT WARREN STREET 

1921 






Copyright, 1905, 

BY 

C. N. SCHMALL 
Copyright, 1921, 

BY 

C. N. SCHMALL 



OF 

BRAUNWORTH & CO. 

BOOK MANUFACTURERS 

BROOKLYN, NTY. 



AUG 18 1921 
©C..A622461 

I 



PREFACE TO THE SECOND EDITION 



In this edition, several proofs and propositions pre- 
viously omitted have been added to the Appendix in the 
form of notes, together with references to the related sec- 
tions of the text. 

The favorable reception which the book has met with 
in England is gratifying to the author, and it is hoped that 
the present edition will prove still more acceptable to the 
class of students who have found it helpful. 

The author is also pleased to observe that the method 
of arrangement, mode of treatment and choice of topics 
herein adopted has been followed by several later writers 
on the subject. Charles N. Schmall. 

New York, 
May, 1921. 



PREFACE TO THE FIRST EDITION 



This work was designed as a text-book for the use of Amer- 
ican colleges and scientific schools. It furnishes a course of 
moderate scope in Analytical Geometry, well adapted to the 
needs of these institutions, and supplies an introduction to a 
more advanced course. Special pains have been taken to give 
an easy and gradual development of the subject, and to make 
the book as attractive and interesting to the beginner as the 
nature of Analytical Geometry would allow. The matter con- 
tained is of an elementary character and can be mastered in a 
few months. 

The treatment of every topic involved is simple, direct, and 
straightforward. Method has been exalted over matter in 
the effort to inculcate a thorough conception of the spirit of 
the subject. The object of Analytical Geometry is not so 
much to derive the properties of the curves investigated as to 
teach a new method of research ; and the student who acquires 
an insight into the method accomplishes far more than he who 
memorizes all the known properties of the conies. 

Each section or article of the book, while serving as a link 
in the chain of arguments, is yet a unit in itself, so that the 
teacher may omit various sections at his discretion without 
impairing the continuity of the work. Throughout the book 
there is a variety of method calculated to relieve the monot- 
ony sometimes found in similar works. Hackneyed and arti- 
ficial devices have been scrupulously avoided, and all proofs 
given are rigorous. Care has been taken to eliminate anything 



vi PREFACE 

vague or abstruse. No means have been neglected in the en. 
deavor to make this a model text-book. 

The vital difference between this book and similar works is 
that it does not attempt to perform the duty of both teacher 
and text-book at the same time. Many recent books contain 
too much in the way of spoon-feeding, and leave little or noth- 
ing for the teacher. Such works, in general, are a failure 
pedagogically. 

The present work presupposes a serviceable knowledge of 
Elementary Algebra, Geometry, and Trigonometry, and a slight 
familiarity with the elements of the determinant notation. 
The elementary properties of determinants are now proved in 
most books on algebra, and it is not expecting too much of the 
pupil to be able to use this instrument. Its advantages are 
well known. The close relations among the above-named 
branches of mathematics are made very apparent in this book. 
Sections involving determinants may be conveniently omitted 
if necessary. 

The exercises in the book have been gathered from* various 
sources, but a great many of them are original. They have 
all been worked out by the author before insertion, in order to 
test their adaptability, and though some of them may tax the 
skill of the best students, none are too difficult for a scholar 
of average ability. The figures are all new and were made 
by the author particularly for this work. 

No specific acknowledgments to other books are given, for 
although a few foreign works were consulted, the matter drawn 
upon is largely the common property of all books on this sub- 
ject, and probably of all mankind. 

Among the features of the book, the following are particu- 
larly noteworthy : — 

(1) The proof of the formulae for the area of a triangle and 
a polygon in terms of the co-ordinates of the vertices. 



PREFACE v ii 

(2) The development of the fundamental relation between 
an equation and its locus, and vice versa. 

(3) The use of the determinant wherever profitable to abbre- 
viate the corresponding operations. 

(4) The treatment of the equation of the second degree 
representing two straight lines. 

(5) The treatment of the more elegant properties of circles, 
viz., coaxial circles, centers and circles of similitude, angle of 
intersection, etc. 

(6) The development of the properties of the conic sections 
Discussion of the ellipse and hyperbola by means of the 
eccentric angle, etc. 

(7) The chapter on the elementary properties of confocal 
comes. 

(8) The numerous exercises. Among them are some of the 
most elegant properties of the conies. 

Other minor details, too numerous to mention here, are evi- 
dent on inspecting the book. 

The author desires to acknowledge his gratitude to his 
publishers for their generous and hearty co-operation: and to 
Messrs. F. H. Gilson Co., of the Stanhope Press, Boston for 
the care they have taken to make the typography attractive. 
Special thanks are due to Professor Oren Root of Hamilton 
College for valuable suggestions, criticisms, and corrections, 
after reading the proofs. Dr. Thomas F. Nichols of the 
same institution also viewed the proof sheets and offered 
helpful suggestions. 

Great precautions were taken to guard against the infiltra- 
tion of errors, but it is likely that the effort has not been 
entirely successful. Any corrections will be gratefully 
received. J 

», v C. N. SCHMALL. 

New York, Jan. 2, 1905. 



TABLE OF CONTENTS 



PAET I 



PLANE GEOMETRY 
Chapteb Page 

I. Co-ordinates. The Point 1 

II. Locus of an Equation 26 

III. Equation of a Locus 41 

IV. The Straight Line 49 

V. The Circle 86 

VI. Transformation of Co-ordinates 124 

VII. The Parabola 132 

VIII. The Ellipse 158 

IX. The Hyperbola 194 

X. Confocal Conics 219 

XI. General Equation of the Second Degree 230 

XII. Higher Plane Curves 241 

PART II 

SOLID GEOMETRY 

I. The Point 275 

II. The Plane 285 

III. The Straight Line 291 

IV. Surfaces. Surfaces of Revolution 298 

Appendix 312 



Part I 

PLANE GEOMETRY 



ANALYTICAL GEOMETKY 



CHAPTER I 



CO-ORDINATES. THE POINT 



1. Application of Algebra to Geometry. — Let the student 
describe a circle of radius, say 5 inches, on an}' part of the 
board, and beneath it write the equation x 2 + y 2 = 25. We 
shall now ask the beginner whether he can perceive any pos- 
sible relation between these two elements (the equation and 
the figure). He will pon- 
der, and tax his imagina- 
tion in a fruitless endeavor 
to associate them, and will 
finally reply in the nega- 
tive.* Now let him draw 
through the center two 
lines X'X,Y'Y, _L to each 
other. [x 2 + y 2 = 25.] 
From any point P on the 
circumference, drop the 
Js PA, PB to X'X, and 
Y'Y respectively. The beginner will now notice that the 
diagonal of the rectangle formed is constant and always 




Fig. i. 



*He will undoubtedly have a little prejudice in favor of the existence 
of some relation though he may be unable to describe it. This manner 
of introducing the study to the beginner will arouse his interest and 
curiosity and produce good results. 



2 ANALYTICAL GEOMETRY 

equal to 5, no matter at which point of the circumference P 
is. Let PB = OA, be denoted by x, and PA by y. Then, 
for any point P on the circumference we have the relation 
x 2 -\- y 2 — (radius) 2 = 25. The beginner will now see that it is 
possible to represent a circle by an equation as above. It is 
the object of Analytical Geometry to show that geometrical 
figures (lines, circles, etc.) may be represented by appropriate 
equations, and that their relations and properties can be con- 
veniently investigated through the agency of these equations.* 



2. Co-ordinates. — For convenience, all points in a plane 
are referred to a pair of intersecting straight lines, generally 
perpendicular to each other, as in the foregoing illustration. 

These lines are termed the axes of co-ordinates or simply 
the axes. A point is completely determined by the lengths of 
the two lines drawn through it parallel to the two axes 
respectively ; or when the axes are rectangular, i. e. JL to each 
other, a point is known when its distances from the axes 
are given. These distances are known as the co-ordinates of 
the point. 

Thus, in Fig. 2a, the point P 
is determined if its distances PM, 
PN, from the axes are known. In 
Fig. 2b, it is determined if the lines 
PN, PM, drawn parallel to the axes, 
and the angle <£ between the axes, 
are given ; X'X is called the axis of 
abscissas, or the x-axis ; Y'Y, the 
axis of ordinates, or y-axis. The 
line PN", measured parallel to the 



N 



II 



X L 



^90' 



III 



M 



IV 



Y 

Fig. 2a. 



* We shall not attempt to give at present a full definition of the subject. 
"The satisfactory definition of any science is often one of the latest 
and most difficult achievements of that science." — Ladd's Psychology. 



CO-ORDINATES. THE POINT 3 

cc-axis, is the abscissa or x of the point P. The line PM, 
measured parallel to the y-axis, is the ordinate or y of P. The 
point 0, where the axes meet, is called the origin. The axes 
divide the plane into four compartments called quadrants. 
The quadrants are designated by numbers, as in Fig. 2a. A 




H5+10?- 



(-x t -y)P 



(-Hc,+lO 



P(+x,-y) 



Fig. 2b. 



Fig. 2c. 



point is designated by its co-ordinates, abscissa first, ordinate 
next ; e. g., the point (x, y), the point (h, k), point (a, b), etc. 

3. Convention of Signs. — If distances measured in one 
direction on a straight line are defined as positive, those in 



Fig. 3. 



the opposite direction are negative. Thus if A and B are two 
points, AB means the distance between the two points and also 
implies that it is measured to the right from A to B. Thus 
AB = - BA, AC = - CA, AC + CB = AB, - BC - CA = 
— BA = AB. This convention of signs is arbitrary, and is 
applied to the four quadrants in the following manner : Dis- 



4 ANALYTICAL GEOMETRY 

tances measured along the #-axis, or parallel to it, from Y'Y to 
the right, are defined as positive ; to the left, negative. Dis- 
tances measured along the ?/-axis, or parallel to it, are positive 
when measured upward and negative when measured down- 
ward. Thus a point (x, y) whose abscissa is x, and ordinate y, 
may have four positions depending upon the signs of its co- 
ordinates. This is clearly shown in Fig. 2c. Hence, the 
signs of the co-ordinates must also be given to fix a point. 
The co-ordinates of the origin are obviously (0, 0). The co- 
ordinates of a point are generally taken according to some 
convenient unit of length.* 

4. Plotting Points. — To plot a point is to locate it. Thus, 
to plot the point (9, 2) we measure 9 units of length along OX 
to the right; then two units upward, parallel to OY. Sim- 
ilarly, the point (—3, — 5) is found by measuring 3 units 
along OX to the left and then 5 units downward, parallel to 
OY'. 

EXERCISES. 



(-5.3) 



(-3-5) 



a, -4) 



Y' 

Fig. 4. 



1. (1) Plot the following 
points : 

..(9,2) (3, -2), (6,-9), (-2,4), 

(-5,-3), (4,6) (A, k). 
■J-J 

(2) The following points : 
(7,4), (0,1), (3,0), (-2,0), 



-I), («,-¥)■ 

Note. — Take ±- as the 
unit. 

2. Construct the A whose vertices are the points (0,0), (3,0), (5,9). 



The teacher should early explain the use of " co-ordinate paper." 



CO-ORDINATES. THE POINT 5 

3. Construct the polygon whose vertices are the points (0, — 2), 

(1,-1), (-1,-1), (-1,0), (2,0), (1,1). 

4. Two vertices of an equilateral A are the points (<z,o), ( — a,o); find 
the co-ordinates of the third vertex. Ans. (o, aVS). 

5. What is the direction, with respect to thex-axis, of the line joining 
the points (2, 3), ( - 6, 3; ; the points (4, 6), (4^_- 7) ; the points (2, 4), 
(6, 8) ; the points (0, 0), (2, 2) ; the points (3 V3, 3), (0, 0) ? 

6 What kind of quadrilateral is formed for the points (0, 0), (a, o), 

(a, 6), (o, b). 

7. The diagonals of a square whose center is at the origin coincide 
with the axes. If its side is a, find the co-ordinates of its vertices. 

8 Two vertices of a square are (a, o), (6, 0) where 6>a; find the 
remaining vertices. 

9. What kind of a quadrilateral is formed by the points (6, 0), (7, 3) 
(2,0), (3,3). 

Note. — Examine the lengths of the sides. 

10. Find the length of the line from the origin to the point (3.4). 



rectangular co-ordinates we 
We can, 



5. Polar Co-ordinates. — In 

determined a point by its distances from the axes, 
however, fix a point by polar co- 
ordinates, in which we take account 
of its distance from a given fixed 
point, called the pole, and its di- 
rection from a fixed line called 
the polar axis or initial line. 
Thus, let O be the pole, and OX 
the initial line ; now let OX re- 
volve about O until it passes 
through P. Then the point P is 
known if the angle 6 and the dis- 
tance OP, called the radius vector 
are known. These together con- Fig , 5 




6 



ANALYTICAL GEOMETRY 



stitute the polar co-ordinates of the point P. The radius vec- 
tor is usually denoted by the Greek letter p, and the puint P 

is designated in polar co-ordi- 
nates by (p, 6). The vectorial 
angle is positive or negative 
according as we suppose the 
line OP to revolve in the 
counter-clockwise or clock- 
wise direction. The radius 
vector p is positive if meas- 
ured from along the re- 
volving line OP, and negative 
if measured in the opposite 
direction, along OP'. Thus the points (- 3, 180°), (2, 90°), 
(3, - 120°), (3, 30°) (4, 180°) are represented by A, B, C, D, and 
E respectively in the figure. 




Fig. 6. 



EXERCISES. 

1. (1) Plot the points (4, 60°), (3, - 180°), (60°, - 2). 

(2) The points, (5, - 300°), (3, 300°), (1, - 30°) (- 2, - 60°). 

2. Plot the points (3, 60°), (- 3, 240°), (3, - 300°) (- 3, -120°). 

3. What line is represented by = 30° ? What is represented by 
P = 2? 

4. Plot the points (4, 150°), (- 5, - 170°), (- 1, 330°). 

5. Plot the points (-a, 45°), (180°, 0), (0, 30°), (-2, 0°), (-a, -30°), 
(3a, -90°), (2a, tan~i2), (a, tan-i i), (&, tan-ii), (30°, -3), (-3, 0°), 
240°, 5). 

Note. — It is immaterial whether the vectorial angle, or the radius 
vector is read first. 



CO-ORDINATES. THE POINT 



6. Transformation from Rectangular to Polar Co-ordinates, 
and Vice Versa. 

Case I. When the polar axis co- 
incides with the x-axis and the pole 
is at the origin. Let P be the point 
(x, y) in rectangular co-ordinates 
and (p, 0) in polar co-ordinates. 

Then from rt. A POQ, 



*j p 




^ 


V 




Ae , 


I 





C v ' 

X 


Q 



Fig. 7. 

[x and y in terms of p, 



X= p COS V 

7 — p sin 

Case II. Pole at point (a, b), polar axis parallel to sc-axis. 

P 




This gives 



x = a - -\- p cos 
y = b -f- p sin 




Case III. Pole at ori- 
gin ; polar axis makes an 
angle <f> with x-axis. 

Here we sret 



p cos (0 + <f>), 
p sin (9 + <p). 



ANALYTICAL GEOMETRY 



Y 




^a_ 

r 


p 





a 


Q 



Fig. 10. 

Case IV. If in Case III the pole is at point (a, h), we 

get: — 

x = a 4- p cos (0 -h <£), 
y = ft + p sin (0 + <£). 



Reversing the process, we get 



Case I. P = V*- 2 + jr 2 , tan0 = -, or = tan" 1 -• 

X X 

Case II. p = V(z - a) 2 + (y - bf, = tan" 1 ^A 

Case III. p = \lx> + ?/ 2 , = tan" 1 ,?> 
Case IV. By the student. 



-<f>. 



7. Radian or Circular Measure of an Angle. — In Higher 
Mathematics an angle is measured by the quotient of its arc 
by its radius. The unit is the radian, which is an angle whose 

arc is equal to its radius. Hence, any angle = ( — =-= — ) radians. 

.-. 360° = 27rr-7-r = 27r radians, or simply 2 tt, 
.'. 180° = 7r radians, 



CO-ORDINATES. THE POINT 



whence 1 radian = 



180 c 



180° 
3.1416 



= 57° 17' 45" nearly. 



EXERCISES. 

(1) Express in radians, these angles : 

1°, 30° r 45°, 60°, 90°, 180°, 270' 
Ans 

(2) Express in radians : 70°, 80°, 220°, 135°, etc 

8. The distance between two points. 

Case I. Rectangular co-ordi- 
nates. 

A(x u y t ) and B (x 2 , y 2 ) 

are the two given points. 

Draw the ordinates AE, BD. 

Draw AC J_ BD. 

Then AB 2 = AC 2 + BC 2 , 

. ■ . d = V(^ 2 - *i) 2 + (7* - «^i) 2 - # 
Case II. Oblique axes. 
Y, 



7T 7T 7T 7T 7T O 

180' 6' V 3' 2' *' 2 f ' 



(^2.2/2.) 





F/fir. 11. 



B (x^Vt) 



Fig 12. 



* The beginuer should take other cases where the points lie in different 
quadrants and convince himself that these formulae are general. 



10 ANALYTICAL GEOMETRY 

Draw the ordiuates AE, BD, and AC II OX. 
Then AB 2 = AC 2 -fIl 2 -2- AC- BC- cos ACB, 
.'.d= V(x 2 - x,f + (y 2 - Vl f + 2 (x 2 - x x ) (y 2 - Vl ) cos <f>. 
If one point, say A, is at the origin, we have : — 



Case I. d= V x 2 2 + y 2 2 . 

Case II. d = V# 2 2 + y£ + 2 x i \J<l cos <A- 
Case III. Polar co-ordinates. 
A (p ly 6j) and B (p 2 , 2 ) are the given points. 
Then AB 2 = OA 2 + OB 2 - 2- OA- OB- cos (0 2 - 0,). 
B (PtM * 



A (Pi,6i) 




Fig. 13. 

.'.<? = V tf + P2 *-2 Pl P2 cos (0 2 - ej* 

EXERCISES. 

1 Find the distance between the points (4, — 5), (— 3, 6). 

d = V [4-(- 3)] 2 +[-5-(+"6f? 
= V7 2 + ll 2 
= >/l70. 
2. Find a point equidistant from the three points (8, — 2), (3, 5), 
(3, - 4). 

Suggestion. — Let (x, y) be the point. 

Then (x - 8) 2 + (y + 2) 2 = (x -3) 2 + (y - 5) 2 = (x - 3) 2 + (y +4) 2 , 
whence x and y can be found. 



* The most important formulae in this work are printed in bold-face 
type. 



CO-ORDINATES. THE POINT 11 

3. Find the condition that the point (h, k) may be at a distance of 
five from the point (7, 9). 

Suggestion. — (5) 2 = (h - I) 2 + (k - 9) 2 . 

4. Express the condition that (A, k) may be equidistant from the points 
(4,5), (-6,8). 

5. Find the length from (2,- 3) to (4, 7). 

6. Show that the points (1, 2), (1, 6), Os/12 + 1,4), are the vertices of 
an equilateral A whose side equals 4. 

7. Three vertices of a parallelogram are (1, 0), (4, 3), (1, 2). Find the 
fourth vertex. 

8. Plot the points (- 3, - tt), (7, - 135°), (5, f w), (0, - § tt). 

9. Find the distance from origin to (3, 4). 

10. Find the distance from (1, 2), (— 5, 4) axes inclined at 60°. 

11. Distance between [3, _), f — — , 5 ]■ 

12. Distance from origin to (— 9, — 6). 

13. Distance between (—4, 6) and (2, 3). 

14. Find the sides of the A whose vertices are (2, 3), (4, — 5), 
(-3,-6). 

15. Express that (a, b) is at a distance of 15 from (2, — 5). 

16. Express that (x, y) is equidistant from (o, b), (a, o). 

17. Distance from origin to (2, 3), (- 2, - 3), (- 2, 3), is Vl3. 

18. Show that the sides of the A ( — 1, - 2), (1, 2), (2, - 3), are re- 
spectively V20^ Vio; V2G. 

19. Show that the A (1, 1), (—1, - 1), (- VW, VS) is equilateral. 

20. Express that (x, y) is equidistant from (— 1, 1) (2, 3). 

21. Show that these points form a square: — (4, 3), (2, 1), (0, 3), (2, 5). 

22. Plot the point { 5, 2 k w + - J, where k is any integer. 

23. Transform these points to polar co-ordinates : — (1, 1), (— 1, 2), 

(-3,3) (-4, -4). 



Ans. (V2\-J , (V5, -tan-^fsV^Trj, ( 4>/2> £ * j • 

24. Transform these points to rectangular co-ordinates — 

( 3 . 4 -M 3 -l)(- 3 -10-(- 3 --f)- 

-•(l^l^)(l,-|vs)(|,-|^(-|v,,|v,} 

25. Find the distance from the origin to (— 2, — 4) axes at 60 4 ". 



12 



ANALYTICAL GEOMETRY 



DIVISION OP A LINE. 

9. To divide the line joining two points (x x , y x ), (x 2 , y 2 ) in a 
given ratio m : n. 




Fig. 14. 

A and B are the given points, C, (x, y) is the required 
point. Draw the ordinates as in the figure. 

AE_ AC _m 

CD ~~ CB ~n 



Now by similar A, 



(i) 



Id 




Iu like manner, 



Fig. 15. 

EC _m 

DB ~ n 

Substituting co-ordinates in (1) and (2) we get, 

x — x 1 m V — V\ m 



x — x n 



y 9 — y n 



(2) 



CO-ORDINATES. THE POINT 



13 



whence 



mx.-, 4- nx. 



y= 



my 2 + ny 1 



221 -f- 22 222 -f- 22 

Again, If AB be divided externally at C, so that 

AC : BC : : m : n, 

Then AC : CB : : m : - rc [since BC = - CB]. 

Hence, by changing n to — n above, we get the required for- 
mulae for external division, viz. : 



x — 



7 = 



221 — 21 

m\ - ny x 



221—21 



10. To bisect the line joining two given points (x x , y^, 
fa, 2/ 2 ). 

T. 




Bfa.y,) 



Let C be the required point of bisection. 
Draw the ordinates and lis as in figure. 
Now AC = CB [by construction]. 

.-. AE = CD, EC = DB. 

y -Vx = y<i -y • • • 



The beginner should prove the generality of these formulae. 



(i) 

(2) 



14 



ANALYTICAL GEOMETRY 



Then from (1) and (2) 



*1 + *2 1 



7 = 



2 

Jx + Ji 



which are the co-ordi- 
> nates of the point of 
bisection. 



Note. —These formulae can be obtained from the preceding article by 
putting m = n = 1. 



,0*i, tfi) 




B&2,y 2 ) 



EXERCISES. 

1. Show that the diagonals of a parallelogram bisect each other. 

Suggestion. — Take two adjacent sides as axes. Find the mid -points 
for both diagonals, which will be identical. 

2. Show that the line joining the mid-points of two sides of a tri- 
angle is equal to half the third side. 

3. In the quadrilateral ABCD show 
that the lines joining the mid-points of 
the opposite sides, and the line joining 
the mid-points of the diagonals meet in 
the point P whose co-ordinates are 

x = I l>i + X 2 + x 3 + x 4~], 

y = \[y\ + V2 + ys + yi\. 

4. Show that the co-ordinates of the 
center of gravity of the triangle {x 1 , yj, 
(x 2 , y 2 ), (x,, y 3 ) are, 

x = [i (x, + x 2 + x 3 )] , 

y = [\ (Vi + vi'+ y 3 )lx- 

5. Find the lengths of the medians of the A whose vertices are 
(-1,-1), (2,3), (0,2). 

6. Show that the mid-point of the hypothenuse of a right A is equidis- 
tant from the vertices. 

Suggestion. — Take the legs of the A as axes. Find co-ordinates of 
mid-point of hypothenuse, etc. 

7. Find the point which divides the line joining ( — 4, 5) and (11,-4). 
in the ratio of 1 : 2. Ans. (1,2). 



c (*3,y 3 ) 



Fig. 18. 



CO-ORDINATES. THE POINT 



15 



8. Three vertices of a square are (1, 9), (5, 6), (2, 2), find the fourth 
vertex. Ans. ( — 2, 5). 

9. The mid-point of a line is (6, 4) ; one end is (5, 7), find the other 
end. Ans. (7, 1). 

10. Find the point of bisection between (2a, 26), (h, k). 

11. Find the point half-way between (5, 8), ( — 3, — 6). 

12. Find the point which divides the distance between (2, 4), (5, — 9) 
in the ratio of 3:4. 

13. Find the points of trisection between the origin and (8, 10). 

14. Two points are A, (4, 1) and B, (9, 8). Find C so that (1) BC = 
AB, (2) BC = 2 AB. 

Note. — A, B, C, must lie on one line. 



11. Area of the triangle (x, , ¥l ), (x 2 , y 2 ), (x z , y s ). 

Y 

C(x 3 ,y 3 ) 




Fig. 19. 

ABC is the given A. The directions of the various lines in 
the figure are easily seen. 
Now, A ABC = A ABG + A ACG + A BCG, 

= |D FDBG + h l3 HGCK + |0 CGBE, 

= I [ O ADEK - E3 AFGH], 

= i [AD x AK - AF x AH], 

== I [ (x. 2 - x,) (// 3 - 1/,) - (x 3 - x x ) (?/ 2 - y x )], 

= h [ X l (i/2 - I/s) ~ l/l ( X 2 - X s) + X 2 VZ - X Z !h], 



16 



ANALYTICAL GEOMETRY 



Cor. 1. If one vertex, say A, is at the origin, the area becomes 

1 

x 2 2/2 1 = H x 2Vs -x z y 2 ). 
Xsys 1 

Cor. 2. If the three points A,B,C, lie on a straight line, the area of A 
ABC is zero. Hence the condition that three points (xi, y{), (05 2 , yi), 
(%s - yz) may be collinear, is 

Xiyi 1 
x 2 y 2 1 =0. 
x s yz 1 

Cor. 3. If the axes are obliqne at an angle 0, we have 
A ABC = \ [ADEK - AFGH], 

= \ [AD x AK sin — AF x AH sin 0], 
X\V\ 1 
.-. A ABC = i sin x 2 y 2 1 
£3 2/3 1 

12. Area of the A when the polar co-ordinates of its ver- 
tices are given. 

A 




Fig. 20. 



A,B,C are the vertices, denoted respectively by ( Pl , 6^, 
(Pi > ^\ (Psj 0»)- 



* The beginner should establish the generality of this formula. 



2. Show that the area of a A = ± I 



CO-ORDINATES. THE POINT 17 

A ABC = A OAB + A OBC - A OAC. 
Area A OAB = \ OA • OB ■ sin AOB, 
= h Pi P2 sin ($ 1 - 2 ). 
AOBC = ip 2 p 3 sm(d 2 -O s ). 
A OAC = i PsPl sin (0,-0,), 

= - i Ps Pi s in (Os - 00, 
.-. A ABC = J [p lPa sin (0 X - 2 ) + P2/ o 3 sin (0 a - 6 8 ) 
+ p 3 Pl sin (0 8 - 00]- 
Note. — The order of the suffixes is cyclic. 

EXERCISES. 

1. Find area of A (0, 0), (2, 3), (4, - 5). Ans. 11. 

x 2 ?/ 2 1 

XsVs 1 

the upper or lower sign being used according as we traverse the periphery 
of the A in the counter-clockwise or clockwise direction. 

3. Find the area of the A (1, 2), ( - 2, 3), (5, 6). Ans. - 16. Con- 
struct the A and by Ex. 2, account for the minus sign. 

4. Find the areas of these A ; 

(1) (8,3), (-2,3), (4, -5). Ans. 40. 

(2) (2,-5), (2,8), (-2,-5). Ans. 26. 

(3) (a, 6), (6, a), (c, c). Ans. \ (a - b) (a + b - 2c). 

(4) of the quadrilateral (0, 0), (5, 0), (9, 11), (0,3). Ans. 41. 

5. Find the area of the A (11, 9), (6, - 2), ( - 5, 3). Ans. 73. 

6. G is the centroid of A ABC ; O is any other point ; prove, 

OA 2 + OB 2 + OC 2 = GA 2 + GB 2 + GC 2 + 3 GO 2 . 

7. Three vertices of a square are (0,-1), (2,1), (0,3); find the 
fourth vertex. Ans. ( — 2, 1). 

8. Show that these three points are collinear : 

(a, 6), (6, a), (3a - 2 6, 3 6 -2 a). 

9. Find the area of the A (o, o), (a, o), (o, 6). 

10. If the angle between the axes is 30°, find the perimeter of A (2, 2), 

(-7,-1), (-1,5). _____ 

Ans. 3 VlO +3 -s/3 + 6 V2+ VS + 3 V2 - VS- 



18 



ANALYTICAL GEOMETRY 



EXERCISES FOR ADVANCED STUDENTS. 

1, Show that the condition that the point (x, y) may lie within the A 
0*4 » V\)i (#2, 2/2), (X3, 2/3) is that the following three determinants shall 
have the same sign : 



x y 1 




x y 1 




x y 1 


xi 2/1 1 


, 


x 2 2/2 1 


, 


£32/3 1 


x 2 2/2 1 




X3 J/3 1 




X12/1 1 



Suggestion. — Take the point inside, join it to the three vertices. 
Then area of given A = sum of three determinants. Now take the point 
outside, obtain area of A. Compare the two results, etc. 

2. A, B, C, are collinear points ; P is any fourth point on this line. 
Prove : 

PA 2 • BC + PB 2 • CA + PC 2 • AB + AB • BC • CA = 0. 

3. Find area of A (axi 2 , 2 ax x ), (ax 2 2 , 2 0X2), (ax 3 2 , 2 ax s ). 

Ans. a 2 (x t — x 2 ) (£2 — x 3 ) (x 3 — x{). 

4. Prove Ex. 2 if P is not on the line ABC. 

5. AB, AC are the adjacent sides of a £? ABDC ; AD a diagonal. If 
the sign of an area is positive or negative according as its periphery is 
gone over in the counter-clockwise or clockwise direction, and P is any 
point in the plane of the ^, 



Prove 



A PAD = A PAC + A PAB. 



6. is the pole, A and B are the points (p t , 6 X ) and (p2 2 ) respec- 
tively. Find the polar co-ordinates of the point of intersection of AB with 

the bisector of the angle AOB. 2 o, o» 

s Ans. = - PlP2 cos I- (02 _ e.) 

Pl + P2 

Suggestion. — Employ the theorem of elementary geometry concern- 
ing the bisector of an angle and the proportionality of the segments of 
the base to the including sides, etc. 

7. P is the mid-point of the base BC of the A ABC. Prove analytic- 
ally that 

AB 2 + AC* = 2 PA 2 + 2 PB 2 



CO-ORDINATES. THE POINT 



19 



13. Digression on Elementary Geometry. — The analytic 
method of investigation, i. e., referring a figure to two fixed 
intersecting lines as co-ordinate axes, and studying its prop- 
erties by Algebraic Analysis, was first introduced by 
Descartes in his Geometrie in 1637. Before his time, how- 
ever, algebra had already been applied to the solution of 
geometrical problems somewhat after the manner of the fol- 
lowing examples. * 

1. Given OA = <x, OB = 6, OC = c, the three lines drawn from the 
center O of an inscribed © to the vertices of the A ABC ; to find the 
radius OF of the 0. 

C 




Fig. 21. 

On CO produced drop the _L BD from B. 
Now A BOD and AOF are similar. 



Let OF = x. 







Reason - 


' Z l + Z 2 + Z s = 90°, 

but zi + Z 2 = Z 4 - 

.-. Z3 + Z 4 = 90°, 

but Z+ + Z$ = 90°. 

1 ••• Z3 = Z5- 






.*. a : x: : 6 : OD, whence OD = — , 

a, 






a 2 


and 


BC 2 = 


JB0 2 + C( 


) 2 bcx 
) 2 + 2- CO- ODJ -& 2 + c 2 + a 



* The teacher should endeavor to dispel the erroneous impression, 
prevalent among beginners, that Analytics is a mere application of alge- 
bra to Geometry. 



20 



ANALYTICAL GEOMETRY 



Again, by similar A BC : BI) ^ OC : OE , 

ah 2 + ac 2 + 2 bcx a 2 b 2 — b 2 x ? 



i.e. 

whence 

or 



: : & : X 2 

a a' 

ax 2 \ab 2 + ac 2 + 2 6cxJ = 6 2 c 2 J a 2 - x 2 \ : 

, ab , ac 5c ) a&c 
*' + *- ^ + 25 + 2li --a" 1 



whence x = OF, the radius, can be found. 

2. XJiven the perimeter p of a right A and the altitude a on the 
hypothenuse. Required the sides. 

Let the sides be AB = x, BC = y, AC = z. 

c Then x 2 +> 2 = z 2 (1) 

x + y + z = P (2) 

From similar & ABC and ABD, 
z:y::x:a, 




or xy = az . . 

By solution of (1), (2) and (3) 
b 2 



(*> 



z = 



2 (a+b) 



2x = b — c +v(6 - c) 2 - 4 ac- 



2y = b-c -V(6- c) 2 



4 ac- 



3. Given the perimeter p of a right A, and the 3 
±s dropped on the sides from any internal point O. 
Required the sides. 

Let OE = a, OF - 6, OG = c, 

Also AB = x, BC = y, AC = z. 

Then x + y + z = p (1) 



x 2 + y 2 



(2) 



ax by cz _ xy 
~2 + ~2 + ~2 = ~2~' 



or ax + by + cz = xy . . . . (3) 

from (1), (2) and (3) the sides may be found. 




CO-ORDINATES. THE POINT 



21 



4. Through a point P within a to draw a 
3hord so that the parts PR and PQ may have 
a given difference d. 

Let PR = x, then PQ = x + d. 

Draw the diameter AB through P. 

Let PA = a, PB = 6, which are known. 

Then PQ ■ PR = PA ■ PB, 

or (x + d) x = ab, 

2 d 

2 



whence 



-V--® 





5. Given the perimeter of a right 
A, P, and the radius, o, of its in- 
scribed ©. Required the sides. 

Let AB = x, 

BC =y, 

AC =z. 



Then z + y +2 = P 



x 2 + ?/ 2 
ax ay az _ 
2" + T + T _ 



/aP\ _ xy 
U/ 2 



(1) 
(2) 

(3) 



cr Whence 



F/flr. 25. 



2a + 64-V4a 2 - 12 a6 + 6 2 

■- 1 

y and z are similarly found. 

Note. — We might also use the 
relation 2a = x + y — z. 

6. Prove : of all rectangles of a given area, the square has the least 
perimeter. 

Let a, and b, be the sides of any one of the rectangles. 
Let x be the side of the equivalent square. Then x = \/ab. 
Hence we must prove 2 x < a + b, or 2\/ab < a + b. 
Now(a-6) 2 >o, .-. a 2 + 6 2 > 2 ab, 

.■ a 2 + 6 2 + 2 a6 > 4 a6, 
.-. a+ 6>2Va&j 
or 2v / a6<a+6. Q.E.D. 



22 



ANALYTICAL GEOMETRY 



7. To divide a given angle ABC into two parts so that their sines 
may have a given ratio* Take BD : BE in the given ratio. Draw BF 

A, 




D B C 

Fig. 26. 
parallel to DE. 6 and <p are the angles required. By Trigonometry, 
sin : sin <f> : : BD : BE. 

8. So that their tangents may have a given ratio. Take any two lines 
AD, DB in the given ratio. On AB describe a segment to contain an 




A D B 

Fig. 27. 
angle equal to the given angle. Draw DC ± AB. 
required angles. 

9. So that the secants may have a given ratio. 

A 



and <p are th& 




D 

Fig. 28. 

Take BE : BD in given ratio. 
DrawBL_LDE. 

d and are the required angles. 

* The following exercises 7-11 will be found useful sometimes in 
doing more difficult ones. 



CO-ORDINATES. THE POINT 



23 



10. To construct a line equal to the square root of a given line. 

AB is the given line, whose length is 

given according to some definite unit. 

Lay off BC, on AB prolonged, equal 
to one unit. 

On AC as a diameter, describe a 
semicircle. 

Draw BL_LAC. BL is the line re- 
quired. 

For BL 2 = AB x BC = AB x 1, 
.\BL=VAB. 

11. To construct two angles, having given the ratio of their sines and 
the ratio of their tangents. 

R 





Take AD : FD in given ratio of the tangents. 

And AD : ED in given ratio of the sines. 

On AF as diameter describe a semicircle, and with D as center, DE 

as radius, describe another. 

Through their point of intersection H, draw AHR. 

DAR = \p, and DHR = \p' are the required angles. 

Draw DQ ± HR, and RN || DQ. 

DQ 

sin \p DA DA DA 

— — V/ = tT7^ = f^rs = ^te; = given ratio. 

sin \p' DQ DH DE & 

DH 

DQ 
tan^ AQ AQ AD 
Ea^ = BQ = HQ = FD = glVenratl °- 

HQ 

Q.E.F. 



Now 



and 



24 ANALYTICAL GEOMETRY 

EXERCISES. 

1. Bisect a given A by a line ± to the base. 

2. Find the side of the square inscribed in a given A if the base and 
altitude of the latter are given. 

3. Given the area of a rectangle inscribed in a given A, to find the 
sides of the rectangle. 

4. To bisect a given trapezoid by a line parallel to its bases. 

5. Given the base, sum of the two sides, and the median to the base 
of a A, to find the sides. 

6. Bisect a given A by the shortest line possible. 

7. Trisect a given A by lines parallel to the base. 

8. In a A, given two sides and the bisector of their included angle. 
Find the base. 

9. In a right A, given the hypothenuse and the side of the inscribed 
square. Find the legs of the A. 

10. Divide a given line into two parts so that the square on one may 
be equal to twice the square on the other. 

11. Find the area of an equilateral A in terms of its side. 

12. Find the sides of a A if the base, the altitude, and the ratio of the 
sides are given. 

13. Also, when the ratio of the sides and the segments of the base 
made by the altitude are given. 

14. Find the sides of a A, having given the base, altitude, and sum or 
difference of the sides. 

15. Determine a A, having given its three medians. Divide a given 
angle into two parts so that : 

16. Its cosines may have a given ratio. 

17. Its cotangents may have a given ratio. 

18. Its cosecants may have a given ratio. 

19. Find the radius of the inscribed circle of a A in terms of the sides. 

20. Divide a given acute angle into two parts so that the rectangle 
of their sines may be a given quantity. 

21. Also, so that the rectangle of their tangents be given. 

22. Determine a right A, having given the radius of the inscribed 
circle and the side of the inscribed square. 



CO-ORDINATES. THE POINT 25 

23. From a given A cut off an isosceles A equal to one-third of the 
given A. 

24. An equilateral A is inscribed in a given square, having one vertex 
at one corner of the square. Find the side of the A in terms of that of 
the square. 

25. Determine a A having given the base, altitude, and product of the 
sides. 

26. Three equal circles are inscribed in a given circle tangent to each 
other and to the given circle. Find their radii. 

27. Transform a given A into an equilateral A. 

28. In a right A, given the hypothenuse and the difference between 
the two lines drawn from its extremities to the in-center. Find the legs. 

29. Also, if the two medians to the legs are given. 

30. Given the points (3, — 1), (6, 4). Find the point of trisection 
nearest the first. Also find the co-ordinates of a point in the line [pro- 
duced] such that its distance from the second equals the distance 
between the first two points. 

31. Find the distance between the points (0, — 2), (1, — 3). 

32. Given the points (1, 2) and (7, — 13). Find the point of trisection 
nearest the first. 

33. Show that these points form a square, viz., (1, 9), (5, 6), ( — 2, 5), 
and (2, 2). 

34. Show that the points (- 4, 1), (2, 3), ( - 1, 2) are collinear. 

35. Find the area of the A (0, 0), (4, - 5), and (2, 3). 

36. Show that these four points form a parallelogram, viz, (5, 2), 
(-1,-2), (2,1), and (2, -1). 

37. Transform to rectangular co-ordinates the equation 

2= 2 « 2 
9 sin 2 6 ' 

38. Find the distance between the points (4, ^ J , f 2 \/2 , — J . 

39. Express in polar co-ordinates the equation y 2 = 8 a:. 

40. Also the equation x 2 — y 2 — k 2 . 

41. Find the area and sides of the A whose vertices are (a, a — 6), 
(6, b — c), (c, c — a). 

42. Show that the points (0, 0), (4, 70°), (4, 10°) are the vertices of 
an equilateral A. 



CHAPTER II 



THE LOCUS OF AN EQUATION 

14. A single equation in two variables x, y, is in general 
indeterminate and admits of an indefinite number of solu- 
tions, i. e.f we can find an indefinite number of pairs of values 
for x and y which will satisfy it. We shall observe, that if we 
think of each couple of values as representing the co-ordinates 
of a point in a plane, we can obtain from the equation a series 
of points all lying on a definite curve. This curve is called 
the locus or graph of the equation. To obtain an idea of the 
general shape and course of the curve, we have to determine 
several points in it (more or less according to the nature of 
its equation), and to draw a continuous " line " through them. 
To this end we solve the equation with respect to one of the 
variables and assign values to the other arbitrarily, but not 
differing widely from one another. We then find the cor- 
responding values of the first variable. The points thus ob- 
tained are plotted and joined consecutively. The process will 
be better understood from a few examples. 

Ex. 1. Construct the locus of the equation y = 5x. 
The following table of values is readily computed. 



X 


1 


2 


o 


4 





i 

2 


3 

2 


- 1 


-2 






etc.* 


























y 


5 


10 


15 


20 





5 

2 


V 


-5 


-10 









* Later on, the student is recommended to begin evaluating with 
z = 0, etc. 

26 



THE LOCUS OF AN EQUATION 



27 



Plotting these points, we observe that they are ranged on a 
straight line which passes through the origin. 

Note. — When an equation has no constant term, its locus passes 
through the origin. 





here 



Fig. 31. 
Ex. 2. Plot the locus of 6 x + 2y — 3 = 0; 

3 - 6x 



Fig. 32. 



V = 



The following table is readily found. 



X 


i 


2 


3 


1 


3 

2 


5 
2 


-1 


-2 







y 


3 

2 


_§ 


-¥ 





-3 


-6 


9 
2 


V 


3 





etc. 



The locus is a straight line, not passing through the origin. 
In the figure each division is \, according to some chosen unit. 

Note. — The beginner will notice as we go on, that the loci of equa- 
tions of the first degree are invariably straight lines. This will be proved 
rigorously hereafter, but we may assume it to be true now as it will 
facilitate the construction of the graph. Since a straight line is deter- 
mined by any two of its points, the graph of an equation of the first de- 



28 



ANALYTICAL GEOMETRY 



gree may be found with less labor. The two points most quickly secured 
are those where the locus crosses the axes ; to find them, put y = o, x = o 
successively in the equation. If A and B are the points required, the 
distances OA and OB are called the intercepts of the curve. To find the 
points of intersection of any two loci, we regard their equations as 
simultaneous and solve for x and y. 

Ex. 3. Plot the graph of y = 2 x + 3. 



X 





l 


2 


3 


i 

2 


f 


-1 


-2 


_3 

2 




y 


3 


5 


7 


9 


4 


6 


1 


-1 








etc. 



The list could be extended indefinitely. 
The points lie on a straight line, shown in the figure. Each 
division is ^. 

Y 










lg. 33. 








Fig. 34 










Ex. 4. 3x + 2y + 6 = 0. 






X 





i 


2 


3 


1 
2 


1 


-i 


-2 


-3 


- \ 








y 


-3 


9 

2 


-6 


_ 15 

2 


_ 1 5 

4 


-y 


-! 





3 
2 


-! 









etc. 



Ths locus is shown in the figure, where each division is J. 



THE LOCUS OF AN EQUATION 29 

Y 




Fig. 35. 



Ex.5. 3a? -4y- 12 = 0. 






1 


2 


3 


2 


3 

2 


-1 


-2 


- 3 


4 




-3 


~~ t 


3. 


— 1 


21 
8" 


1 5 

~8~ 


1 5 
4 


1 8 
"4 


2 1 

■i 








etc. 



The graph is shown in the figure. Each division is J. 
Ex. 6. Construct the locus of y + x = 0. 

Y 




Fig. 36. 



The student can easily compute the table of values. The 
line passes through the origin, and the second and fourth quad- 
rants, cutting the x-axis at an angle of 45°. 



30 



ANALYTICAL GEOMETRY 



Ex. 7. Construct the graph x — y = 0. 

By the student. 

Ex. 8. Construct the locus of x = 3. 

This curve evidently has its abscissa always equal to 3. It 
is .-. a line J_ to cc-axis and at a distance 3 from the origin to 
the right. 



Fig. 37. 

Ex. 9. Construct the locus of x = — 4. 
By the student. 



X-!- 



G 



Y 

Fig. 38. 
Ex. 10. y - 2 = 0. 

This is evidently a line parallel to the #-axis and at a dis- 
tance 2 above it. 



Ex. 11. y + 5 = 

By the student. 

Ex. 12. Construct the locus of 9x 2 + 25 y 2 = 900. 



y = ± \ V 900 



THE LOCUS OF AN EQUATION 



31 



The following table is readily found. 



± i 


±5.96 


± 2 


±5.87 


± 3 


±5.72 


± 4 


-J- 5.49 


± 5 


±5.19 


± 6 


±4.80 


± 7 


-J- 4.28 


± 8 


-J- 3.60 


± 9 


±2.61 


±10 


±o 





±6 


etc. 


etc. 




Plotting these points accurately and joining them by a 
smooth curve, they are found to lie on an ellipse. 

Ex. 13. Plot the 
curve y- = 5 x, 

y = Vb x. 




1 


±2.23 


2 


±3.16 


3 


±3.87 


4 


±4.47 


5 


±5.00 


6 


±5.47 


7 


±5.91 


etc. 


etc. 



Fig. 40. 

For negative values of x the value of y is imaginary ; hence 
no point of the curve lies to the left of the y-axis. When 
x = oo , ?/ = oo, hence the curve extends to the right without 
limit. It passes through the origin, The curve is called a 

parabola. 



32 ANALYTICAL GEOMETRY 

Ex. 14. Construct x 2 + y 2 = 16. 






±1 


±2 


±3 


±4 










±4 


±8.9 


±3.5 


±2.6 












etc. 



The points are found to be arranged on the circumference 
of a circle whose center is at the origin and whose radius is 
equal to 4. 

r, 




x x 




xy = 4. 
4 

y = z- 



X 





±1 


±2 


±3 


±4 


GO 








y 


00 


±4 


±2 


=Lt 


±1 












etc. 



Plotting these points, we get a curve consisting of two 
branches, one lying in the first quadrant, and the other in the 
third quadrant. The curve is known as the hyperbola. 



Ex. 16. Construct 


x 2 - 3 x - 


. 2y + 1 = 


x 


- 3x+ 1 


y = — 


2~~ 


X 


y 





i 


l 


- i 


2 


- i 


3 


4 


4 


1 


-1 


I 


- 2 


-V- 



THE LOCUS OF AN EQUATION 
F 



33 




Fig. 43. 



Plotting the points, we find the curve shown in the figure. 

x 2 y 2 



Ex. 17. Plot the curve ~ ^ = 1 



X- 




Y' 

Fig, 44. 




By the student. 

The curve is an hyperbola and has the shape given in the 
figure. 



34 ANALYTICAL GEOMETRY 

EXERCISES. 

«■»■ 

Construct the following graphs or loci : 



1. 


y- = 4 x. 




11. 


x 2 + y* = 9. 


2. 


x 2 = 9y. 




12. 


x 2 = 4 y. 


3. 


3 x 2 - 4 ?/ 2 = 


12. 


13. 


x 1 + y 1 = 4. 


4. 


4 x 2 + 3 ?/- = 


10. 


14. 


x 2 + x + y — 1 = 


5. 


y = x 3 - 1. 




15. 


x 3 — 2 x — y = 0. 


6. 


x - y^ + 1. 




16. 


2x + y-5 = 0. 


7. 


y* = 2 - x*. 




17. 


4 xy - 1 = 0. 


8. 


x 3 = 4 ?/. 




18. 


y = 2x 2 + 1. 


9. 


x - 2/ = 3. 




19. 


x 2 - x - 1 = 0. 


0. 


2x -y= 1. 




20. 


x 2 = 6 ?y. 



15. Discussion of Equations. — The discussion of an equa- 
tion is a critical examination made in order to ascertain the 
peculiarities of its locus, such as intercepts, limits of extent, 
symmetry, continuity, etc. The discussion of an equation of 
the first degree, which represents a straight line, consists gen- 
erally in finding its intercepts and its direction with respect 
to the axes. 

Discussion of Ex. 12. — If x = 0, y = -J- 6 ; if y = 0, x = -J- 10. 

These are the intercepts of the curve on the axes. 

If x > 10, y is imaginary. 

If y > 6, x is imaginary. 

Hence x = -J- 10, y = -[-6, are the greatest abscissas and ordiuates re- 
spectively, of the curve. Again, for every value of x, there are two equal 
values of y, but unlike in sign ; hence the curve is symmetrical with re- 
spect to the x-axis. 

Similarly it is symmetrical with respect to the y-axes. It is therefore 
symmetrical with respect to the origin as a center. 

Discussion of Ex. 13 — If x =,0, y = ; hence, curve passes through 
the origin. * 

Negative values of x give imaginary values of y ; hence no point of the 
curve lies to the left of the y-axis. 



THE LOCUS OF AN EQUATION 



35 



If x = oo , y = co ; the curve therefore extends to the right of the y-axis 
without limit. 

Again ; for every value of x there are two equal and opposite values 
of y ; the curve is therefore symmetrical with regard to the cc-axis. 

Ex. 18. Trace the curve, y = sin x. 



x (in 


degrees). 


y 


X 


y 


0° 




0.00 


190° 


- .17 


10° 




.17 


200° 


- .34 


20° 




.34 


210° 


- .50 


30° 




.50 


. . . 


.... 


40° 




.64 




.... 


50° 




.77 






60° 




.87 


270° 


-1.00 


70° 




.94 






80° 




.98 






90° 




1.00 






80° 




0.00 







The curve has the form shown in the figure. 
Y 

v * 




Fig. 45. 



Discussion. — The curve cuts the x-axis at intervals of 180°, since 
at these intervals sin x = 0. It extends without limit on both sides of the 
origin, since an angle may have any magnitude positive or negative. 
The ordinate has its maximum values alternately + 1 and - 1. The 
positive value corresponds to 90°, and the negative value to 270°. They 
recur at intervals of 360°. See Fig. 172, end of book. 



36 



ANALYTICAL GEOMETRY 






Fig. 46. 




Ex. 19. Plot the curve y = 


tan x. 




X 




V 


0° 


0.00 


10° 




.18 


20° 




.36 


30° 




.58 


40° 




.84 


50° 




1.19 


60° 




1.73 


70° 




2.75 


80° 




5.67 


90° 




00 


- 10° 




- .18 


-20° 


' 


- .36 



etc. 



etc. 



Discussion. — The curve cuts the z-axis at intervals of w ; i. e., a 
J- 2 7r, J- 3 7T, etc. It is discontinuous at intervals of w ; i. e., at -j- - 

_|_ - 7r, -1- 5 7T, etc. When x passes through the value |, y changes from 

+ oo to - oo ; and as x passes from £ to \ ir, y passes from - oo through 0, 

L — ■ 

to + oo. As in the case of the sine curve, this curve has an infinite 
number of branches. 



THE LOCUS OF AN EQUATION 



37 



Note. — The trigonometric curves may 
be readily traced by using the line values 
of the functions, thus : 

If \{/ = POA is any angle at the cen- 
ter O, of © with radius equal to unity, 
we have 



sin \f/ = PN. 
cos \p = ON. 
tan \p = AD. 



ctni/' = BM. 
seo/' = OD. 
esc \J/ = OB. 




Fig. 47. 



Trace these curves : 



EXERCISES. 



1. 


y = 3 sin x. 


6. 


y =3 esc x. 


2. 


2 y = ctn x. 


7. 


y = sin x + cos x. 


3. 


y = i cos X. 


8. 


?/ = tan x + 2 ctn x 


4. 


?/ — ctn x. 


9. 


x = sin y. 


5. 


?/ = sec x. 


10. 


2 x = cos ?/ + sin y. 



16. Definitions. — The student is now prepared to give a 
definition of the locus of an equation; it is the " line" or group 
of lines such that the co-ordinates of every point on it satisfy 
the given equation. 

Conversely. — The equation of a locus is the equation satis- 
fied by the co-ordinates of every point on the locus. Or we 
might define a locus as the path of a point which moves under 
a given condition, i. e., that imposed on it by a given equation. 

Note 1. Definitions. — A constant is a quantity or symbol which 
does not change its value throughout the same discussion. A variable 
may have any value. An absolute constant is a numeral or a symbol 
representing a numeral ; e. g., w. = 3.1416, \/5, 16, etc. 

An arbitrary constant is one which is constant in one investigation 
and is generally represented by the first letters of the alphabet, a, &, c, h, 7.\ 
Variables are represented by the last letters of the alphabet, e. g., x, y, z, 
W), etc. 



38 ANALYTICAL GEOMETRY 



Fig. 48. 



Note 2. Definitions. — The left-hand side of 
an equation is called the sinister, the right side the 
dexter. 

Exercise. — Show that the locus of the equa- 
tion y = x 3 has the form given here. 



17. Product of Two or More Equations. — The locus of 
the product of two or more equations whose dexters are zero 
embraces the combined loci of the equations. Thus, the locus 
of 

The product rS 1 = , | . . . . . (1) 

of \ S 2 = I (2) 

(1), (2), (3), I S 3 = J (3) 

is represented by S x ° S 2 • S s = 0, for it evidently contains all 
points on these loci. 

Examples : -*- 

(1) Find the locus of xy = 0. Arts. The axes 

(2) x 2 — y 2 = 0. Ans. The two lines x = y, x = - y 

(3) 4 x 2 - 9 y 2 = 0. Ans. The lines 2 x - 3 y = 0. 

2 x + 3 y = 0. 

(4) x 2 - 6x + 8 = 0. Ans. The lines x - 4 = 0. 

x - 2 =0. 

(5) x 2 + 8xy + 12y 2 = 0. Ans. The lines x + 2y = 0. 

a; + 6 i/ = 0. 

(6) y 3 - x 3 = 0. ^4ns. The curves y - z = 6, 

and y z + xy + x 2 = 0. 

EXERCISES. 

Trace the following curves : 

1. 3x - 4y = - 12. 4. = 0. 

2. x 2 + y 2 = 49. 5. = 60°, 30°, 45°, 120°, etc. 

3. y 2 - x 2 + x 3 = 0. 6. p = 4 sin' 0. 



* Note. — It is obvious that the locus of an equation whose dexter is 
zero, can be more readily traced if the sinister is factorable. 



THE LOCUS OF AN EQUATION 39 

7. y 2 = 9x 2 . 

8. x 2 = 16 y 2 . 

9. x 2 - xy + 1 = 0. 

10. y 2 = 4. 

11. x 2 = 9. 

12. x 2 - y 2 = 6. 

13. y = x + 2. 

14. y 2 = 16 x. 

15. x + y - 9 = 0. 

16. 3x - 5 = 0. 

Find the points of intersection of the following loci : 

26. y 2 = 4x, y = 6. 31. y = - x 3 , x - y = 0. 

27. x + y - 4 = 0, 3x + y=6. 32. x + y = 4, y = 5. 

28. - + | = 1, ? + ^ = 1. 33. x 2 + y 2 = 9, y = 3x - 1. 

29. x 2 + y 2 = 25, x = 2 y. 34. p = 2 cos 0, p cos d = 4. 

30. y 2 = 6 x, 2 x + 3 y = 0. 35. p = 2 a cos 0, tan 5 = 3. 
36. y = sin x, y = 3 cos x. ( p = _]_ 2 V% 



17. 


2x - Sy = 6. 


18. 


xy = l>6. 


19. 


x L + y 2 = 1. 


20. 


,-*,»-\ 


21. 


x 2 -2y 2 = 0. 


22. 


x+ 3y = 9. 


23. 


y — 5x = 4. 


24. 


3x - 22/ = 6. 


25. 


/> = 2a cos 6. 



6 = cos- 1 V2- 

37. Find the intercepts of these loci : 

(1) x 2 + y 2 - 4x - Sy - 5 = 0. 

(2) 2 (x - 2) 2 = y + 1. 

(3) y = x 3 - 3x 2 +14. 

(4) x 2 + y 2 - 8x + 6 y + 12 = 0. 

mS + S-i- 

38. Find where the curves p = 2 sin 5, and p 2 = 4 cos 2 5 cut the 
polar axis. 

39. Find a point on the curve y 2 = 9x, whose abscissa is twice its 
ordinate. 

40. Find the area of the A formed by the three lines, x — y = 0, 
x + 2 y = 0, and — 3 x + 2 y = 5. 

41. Find the distance between the points of intersection of the curves 
x ? + if = 49, 2 x + y = 0. Also, y 2 = 4 x, x = y. 

42. For what values of k will the curves x 2 + y 2 = 4, and 2x + Sy 
— k — not intersect? 

43. When will the curves y 2 = 4x, and y = 2x + k meet in two co- 
incident points? 



40 ANALYTICAL GEOMETRY 

44. Find the points of intersection of the loci, 

4 + 1 ' 4 1 ~ im 

45. Write the equation of the locus passing through the points of 
intersection of the curves x 2 + y 2 = 25, and y — 3x + 4. 

46. Construct the locus of p = 2 6. 

47. Construct p = b 2 cos 2 0, and p = b sin 2 d. 

48. Construct 2 y 2 = 3 x\ y = 2x 3 , x.= y 3 . 

49. Construct x 3 — ?/ 3 = 0, x — x 3 = — 8, x 3 — x= 0. 

50. Construct x 2 - 9 y 2 = 0, 4 x 2 — 25 ?/ 2 = 0. 

51. Find the points of intersection of the curves, 

x 2 y 2 _ x 2 2/ 2 _ 

4~~25 ' 4 + 25 -1 ' 

x 2 t/ 2 

52. Find a point on the curve — — ^- = 1 whose abscissa is 3 times 

Zo lo 

its ordinate. 



CHAPTER III 



THE EQUATION OF A LOCUS 

18. To reverse the process of the preceding chapter, i.e., to 
find the equation of the path or locus of a given point, when 
we are given the conditions under which that point moves, is 
a more difficult and complicated task. We shall endeavor 
to introduce the student to the method by a series of easy 
examples. 

Ex. 1. To find the equation 
of the locus of a point P, which 
moves so that the sum of its dis- 
tances from two fixed points A 
and B is constant and equal to 
2 a. 

Take the line AB as x-axis 
and its _L bisector as t/-axis. Let 
AB = 2 c. Also let P (x, y) be 
the position of the moving point 
at any moment. Then, by the 
condition, we have, 

V(x + c) 2 + y 2 + VV - c) 2 + y 2 
or V{x + c) 7 + y 2 = 2 a — VJx 

Square both sides and simplify : 



I 


f(x.y) 

\ 

\ 

\ 
\ 


A (~c,o) 


(c.o)B 



Fig. 49. 
= 2a 



c) 2 + y 2 . 



Squaring again, 
which sives, 



a \/(x — c-) + y 2 = a 2 — ex. 
a 2 (x — c) 2 + a 2 y 2 = (a 2 — ex) 2 , 
(a 2 — c 2 ) x 2 + a 2 y 2 = a 2 (a 2 — c 2 ). 



Pat a 2 



6 2 , and divide the equation by its dexter, we get 
1. 



a 2 "*" 6 2 



(1) 



41 



42 



ANALYTICAL GEOMETRY 



This equation represents an ellipse, and is the required 
locus. The curve will be discussed in a subsequent chapter. 

Ex. 2. If hi Exercise 1 the difference of the distances is given equal 
to 2 a, find the locus of the point P. 

The work is similar, giving for the required locus, the equation, 



V 



1. 



B(-a,o) 



hD (a,o)C 



This represents an hyperbola which will be discussed later. 

ISx. 3. A, B, and C are any three points not in the same straight 
line. Find the locus of a point P in their plane which moves so that 

PB 2 + PC 2 = 2 PA 2 . 

Let BC = 2 a, and take BC as 
z-axis, and its JL bisector as ?/-axis. 
Let A be the point (h, k). 
Now PB 2 = (x + a) 2 + y 2 ; 
PC 2 = (x - a) 2 + y 2 . 
PA 2 = (x- h) 2 + (y - k) 2 . 
.-. By the condition, 
(x + a) 2 + y 2 + (x - a) 2 + y 2 
- 2 [(x - h) 2 + (ij - k) 2 ] . (1) 
Fig. 50. 2x 2 + 2 y 2 + 2a 2 = 2(x-h) 2 +2ty-k) 2 , 

which reduces to, 
2 xh + 2 yk = h 2 + k 2 - a 2 , 

which is the required locus, a straight line, since its equation is of the 
first degree. The result is the same if the point P is taken in any other 
quadrant than the first, the proper signs of its co-ordinates being ob- 
served. 

Note. — Let the student work this exercise by taking any two ± 
lines as axes, and calling A, B, and C, (x,, t/,),(z,, y 2 ,), and (ar 3 , y 3 ) re- 
spectively. The result will also be of the first degree, but the beginner 
will see the advantage of a careful choice of axes. 

Ex. 3. Find the equation of the locus in Ex. 2 when the three given 
points are collinear. 

By the student. 



THE EQUATION OF A LOCUS 



43 



Ex. 4. Given the base of a A = 2 a, and the difference between the 
squares of its sides = k 2 . Required the locus of its vertex. 

Take the base AB and its ± bisector 
as axes. Let P (x, y) be one position 
of the vertex. 



-2 

L 

PB 2 = (x - a) 2 + y 2 . 
.-. By the condition, 

AP 2 - BP 2 = k 2 , 

or, ( x + a) 2 +y 2 -[(x-ay + y 2 ]=k 2 , 
whence, 4 ax = k 2 . 

k" 



A(-a.o) 



P(x.y) 

' IN 



(a,o)B 



Fig. 51. 



4a 



is the equation of the required locus. 



It is a straight line JL to the rc-axis. 

Ex. 5. A line parallel to the base of a A cuts off a trapezoid. Find 
the locus of th3 point of intersection of its diagonals.* 

Let the sides AB, AC of the 
given A ABC be the axes. Let 
DE be || to BC. Also let AB = a, 
AC = 6, and let AD and AE be 
respectively equal to ka, kb, since 
they are proportional to the sides 
a and b. 

Then the equation of BE is 




(ka.oW 
Fig. 52. 



B(a.o) 



a kb 



x 
ka 



= 1 



L • • (1) 

• • • (2) 
, obtaining 



And the equation of CD is 

Subtract one equation from the other, and divide by (1 — r J 

as a result j- = 0, y = — x which is the required locus, a straight 

line, the median on BC. 



* Note. — The beginner should defer reading this exercise until he 
has read the chapter on the straight line. 



44 



ANALYTICAL GEOMETRY 



Ex. 6. In a A OCD, we are given the base = c, and the sum of the 
two varying sides = m. The altitude CM is produced beyond the vertex 
to P, so that MP = OC always. Find the locus of the point P. 

Take the base OD as x-axis, and a 
± to it at O as ?/-axis. 

Then OC = MP = y, 
CD = m - y. 

Now, 




MC = OC 
And 
MC 2 = CD 2 -MD 2 



OM 



y 



x\ 



y 



Reducing, this gives 2 ex — 2 my + m 2 



(m 
= 0, 



= (m-y) 2 -(c-xY-. 
y? - (c-a0 2 (l). 



which is the required locus, a straight line. Another solution : 
Take the origin at the mid-point of the base. Let the A be ABC. 
Then AC = MP = i/, 

And BC = m — y. 
Now, 

BC 2 = AC 2 4- AB 2 -2AB x AM, 



P(x.y) 



1-4 



or, (m—yy = if + c 2 — 2c 

whence, m 2 — 2 my= —2 ax, 
c m 

y =m X + l' 
which represents a straight line as 
before. 




f(oc.y) 



(a,o) C 



Fig. 55. 



Fig. 51. 

Ex. 7. Find the locus of point P, 
whose distance from a given lins is 
proportional to the square of its dis- 
tance from a given point. 

Take the fixed line AB as ?/-axis, 
and let CO drawn ± to AB from the 
given point C be the ar-axis. 

Put OC = a. Let P (re, y) be any 
position of the moving point. 



THE EQUATION OF A LOCUS 



45 



Then, by the conditions, PA = k PC , where k is a constant, 
or, x = k [(a - x) 2 + y 2 ], 

which represents the required locus. 

Ex. 8. If in Ex. 4, the sum of the squares of the sides = 2 A; 2 , find 
the locus of the vertex. Ans. x 2 + y 2 = k ? — a 2 . 

Ex. 9. Given the base of a A = 2 a and the product of the tangents 
of the base angles = k (a constant), find the locus of the vertex P. 

Take the origin at the mid-point of 
the base. 

y 



Then tan A = 
tan B = 



a + x 

y 



a — x 
then, by the condition, 

y . y 

a + x a — x 
:. y 2 = k (a 2 
»s the required locus. 



k. 



x 2 ) 



T 


\ 
1 \ 
1 \ 
1 \. 


A (-a,o) 


(a.o)B 



Fig. 56. 



Ex. 10. Given the base of a A = 2 a, and the ratio of the squares of 
vts sides, m:n ; find the locus of the vertex. Ans. A circle. 

Ex. 11. Given the base of an isosceles A = 2 a, find the locus of its 
vertex. 

Ex. 12. Given the base of a A = 2a, and the area = c 2 , find the locus 
of its vertex. Ans. A line parallel to the base. 

Ex. 13. Given the base and the ratio of the sines of the base angles, 
find the locus of the vertex. Ans. A circle. 



Note. 

sides. 



Observe that these sines are to each other as the opposite 



Ex. 14. Find the locus of a point such that the sum of the squares of 
its distances from the four corners of a given square is constant, = 8 k 2 . 

Suggestion. — Take the side of the square — 2 a, and the required 
locus is a with its center at the center of the square and a radius 
^ V2 (k 7 - a 2 ). 

Ex. 15. Find its locus if the sum of the squares of its distances from 
the sides of the square is constant. Ans. A circle. 



46 ANALYTICAL GEOMETRY 

Ex. 16. A and B are any two points. Find the locus of the point P 
such that PA = k ■ PB. Ans. (x + a) 2 + y 2 = k 2 { (x - a) 2 + y 2 \ • 

Suggestion. PA 2 = k 2 • PB 2 . 

Ex. 17. A point moves so that the sum of its co-ordinates is always 
equal to 3. Find its locus. Ans. x + y = 3. 

Ex. 18. A point moves so that the sum of the squares of its co-ordinates 
is equal to its abscissa plus 7. Find the locus of its vertex. 

Ans. x 2 + y 2 = x + 7. 

Find the locus of a point which moves: 

Ex. 19. So that the sum of the squares of its co-ordinates is always 
equal to twice their product. Ans. The line x — y = 0. 

Ex. 20. So that the difference of its distance from two _L lines is 
equal to 5 times its distance from their intersection. 

Suggestion. — Take the fixed lines as axes. 

Ex. 21. So that it is always equidistant from the points (2, 4) and 
(3, - 5). 

Ex. 22. So that its ordinate is \ of its abscissa plus 9. 

Ex. 23. So that its ordinate equals its abscissa. 

Ex. 24. So that its distance from (h, k) is 3 times its distance from 
(«, b). 

Ex. 25. So that it is always at a distance k from origin. 

Ex. 26. So that it is always at a distance k from (a, b). 

Ex. 27. A, B, C, and D are four given points. Find the locus of P 
such that A PAB + A PCD = k 2 (a constant). Ans. A straight line. 

Ex. 28. Also so that A PAB - A PCD - k 2 . 

Ex. 29. Polar co-ordinates.* A line revolves around a fixed point O, 
meeting a given line PM in a point P. Find the locus of the point Q in 
the revolving line so that OP • OQ = A; 2 . 

Take O as the pole, and OM _L to PM/as polar axis. Let the point Q 
be (p, 0), and let OM = a. 

Then a = OP cos 6. 



* When a line revolves about a fixed point, and the locus of a point 
on the moving line is desired, it is more convenient to use polar 
co-ordinates. 



THE EQUATION OF A LOCUS 



47 



But, by the condition, 



OP 



k 2 , or OP = 



A; 2 



[equation of the locus]. 
Y 




>(PJ) 



M 



-X 



Fig. 57. 



k 2 
or .'. a = — ■ cos 0, 

P 
ap = k 2 cos 0, 
which represents the required locus. 

Note. — To transform to rectangu- 
lar co-ordinates, this result may be 
written : 

a p 2 = k 2 • p cos 0, 
or a (x 2 + y-) = k 2 x, 

which shows that the locus of the 
point Q is a circle. 

Ex. 30. Trace the locus of the equation p = 2 a cos 0, by assigning 
values to and finding the corresponding values of p. Then transform 
to rectangular co-ordinates. 

EXERCISES. 

1. Given the base and the median drawn from one extremity [in a A], 
find the locus of the vertex. 

Suggestion. — Let b = base, m = median. Let (h, k,) be the other 
extremity of the median, where the base is the x-axis, and the extremity 
is the origin. Ans. A circle, center (— c, o), radius 2 m. 

2. Find the locus of a point which moves so as to divide in a given 
ratio all the lines that can be drawn from a given point to a fixed line. 

3. Find the locus of a middle point of a line of constant length, a, 
which moves so that its ends always touch two fixed ± lines. 

Suggestion. — Take the fixed lines for axes. .4 ns. 4 (x 2 + y 2 ) = a 2 . 

4. A square is so moved that one of its diagonals always has its 
extremities in two fixed ± lines. Show that the second diagonal has 
its extremities [always] in two other fixed ± lines. 

5. One side AB of a A is given in length, and another side AC of 
constant length revolves about the fixed point A. Find the locus of the 
mid-point of the third side. 

Suggestion. — Put AB = a, AC = c. Take AB as axis of x. 



Ans. (x- a -y=(iy-y\ 



48 ANALYTICAL GEOMETRY 

6. A right A is moved so that the extremities of the hypothenuse 
always touch two fixed ± lines. Find the locus of the vertex of the 
right angle. 

7. Two points P and Q are taken in the sides AB and AC of the A 
ABC, such that AP • AQ = BP • CQ. Find the locus of the mid-point 
of PQ. 

Suggestion. — Put AB = a, AC = b. Take AB and AC as axes. If 
(x, y) is the mid-point of PQ, we have AP = 2 x, AQ = 2 y. 

Ans. 4 xy = (a — 2 x) (b — 2 y), a straight line. 

8. A and B are two fixed points. If PA and PB intercept a constant 
length, k, on a given line, find the locus of P. 

Suggestion. — Let AB meet the given line in 0. Take the given line 
and AB as axes, and as the origin. 

Let OA = a, OB = b. Ans. k (a — y) (b — y) = (a — b) xy. 

The base AB of a A is given, where A is ( — o, o), B (a, o). 
Find the locus of the vertex C if, 

9. Cot A + m cot B = k. Ans. (1 — m) x — ky + (1 + m) a = 0. 

10. B = 2 A. Ans. 3 x 2 - if + 2 ax - a 2 = 0. 

11. m • AC 2 + nBC 2 = k 2 . 

Ans. (m + n) (x 2 + y 2 ) + 2 (m — n) ax + (m + n) a 2 — k 2 = 0. 

12. A - B = 0. .4ns. x 2 - y 2 - 2 xy cot <f> - a? = 0. 

13. A straight line of variable length, OP, revolves about O, so that 
OP is equal to the product of a constant k, by the cosine of the angle 
which it makes with the initial line. 

Find the locus of P. Ans. p = k cos G. 



CHAPTER IV 

THE STRAIGHT LINE 

19. The equation of the straight line through two given 
points (x,, y x ) (x 2 , y 2 ). 



A 



(x,y) ^2.2/2) 




Fig. 58. F '9- 59 - 

Let P (x, y) be any point on the line. Draw the ordinates 
and lis as indicated in the figures. 

BF PE 

Then by similar A, p= = y^ • 

. lh - Vv _ V-V\ ) 
' ' x — x x x — x x 



or, 



y - ih hh - ?h\ _ ih - y>. 



X — X-i \Xo x 



which may be written, 



x j 1 

^., Jo 1 



(1) 

(2) 



Both forms should be remembered. 

49 



50 



ANALYTICAL GEOMETRY 



20. Condition for three collinear points. The line through 
the points 



6*2 > 2/ 2 )> 6*8 > V*)> is 



x y 1 

^2 3/2 1 
*3 2/3 1 



0. 



Now, if the point (x lt y x ) lie on this line, we have 

x x y x 1 



^2 I/2 1 
^3 2/3 ! 



-0, 



which is the condition that the three points (x x , y x ), (ar 2 , y 2 ), 
(.x 3 , 2/ 8 ) may be collinear. 

Note. Observe that this result agrees with that of § 11, cor. 2. 

21. Symmetrical form of the equation of a straight line ; 
or the equation in terms of its intercepts, a and b. 





OA = a, OB = b. 
P (x, y) is any point on the line. 

Now, by similar A, ^ = ^ . 



whence, 



a a — x 
a Z> 



a) 



THE STRAIGHT LINE 



51 



This equation may also be found thus : Since the line 
passes through the two points (a, o), (o f b), its equation, by the 
preceding article, is 

x y 1 

a o 1 =0: 

o b 1 



whence, 



x y . _ _ 
- -h 7- = 1, as betore. 
a b 



Query. — What must be the signs of a and b, in order that this line may 
cut Quadrant I? II? III? IV? 

22. Equation in terms of slope and intercept, where the 
slope is tan $, the angle which the line makes with the 
x-axis. 

Intercept OB = b. Let P be 
any point on the line. 

Slope = tan is generally 
represented by the symbol m. 

Then, in the figure, 

PE v-b 






y 




BE 



Fig. 62. 



.-. j = mx 4- b, the required equation. 

Query. — What line is represented by the equation 
x = ny + a? 

23. Equation of a line through 
the origin. 

Let P (x, y) be any point on 
the line. 




Then, 



y 



Fig, 63. 



or, 



• • • (1) 



52 



ANALYTICAL GEOMETRY 



which, is the form of the equation of a line passing through 
the origin. 

Example 1. — Find the equation of a line through the origin and the 
point (1, 2). 

Let y = mx be the required equation (m undetermined). 
Then, since it passes through the point (1, 2), we have, 
2 = m • 1, or m = 2. 
. •. y = 2 x is the required equation. 
Example 2. — Eind equation of line through (0, 5) and making an 
angle of 60° with x-axis. 

Here b = 5, m = tan 60° = V3. 

. •. y = \/S x + 5 is the required line. 
Example 3. — Find the line through (—2, 3) and at 150° to x-axis. 
Let y = mx + 6 be the required equation, 

1 

and since line passes through point ( — 2, 3), we have 

2 



m = tan 150° 



--a 



( - 2) + 6 



v/3 



+ b. 



■. b 



V3 



.'. y= -x+ 3 

V3 



- is equation required. 
3 

Find the equation of the line through the points (2, 3), 



Example 4. 
(5, 7). 

Let y — mx + b be the required equation. 

Then 3 = m • 2 + b (1) 

and 7 = m ■ 5 + b (2) 

From equations (1) and (2) m and b are readily obtained. 

24. Equation in terms of one 
point (x ± , y x ) and slope m. — 

P is any other point (x, y) 
on the line. 

_, PD y - y x 

Then m = =^- = ^ • 

i)A x — x. 



J 




f ^ 




J^- 


— \p 










Fig. 64. 



/■y~ 7 1 = m(JT- 2q) 
is the required equation. 



THE STRAIGHT LINE 



53 



25. Normal Equation of the straight line. — This is the 
equation in terms of p (the J_ distance of the line from the 
origin) and a (the angle 
which this _L makes with 
the ic-axis). 

Z NO A =a, OD = p. 

P (x, y) is any point on 
the line. 

PMisl_OA,PCis_LMN, 
which is parallel to AB. 
ZCMP=a. (Why?) 

OD = p = ON + CP == x cos a + y sin a, 
or, ^"COS a + .rsin a = P (1) 

which is the required equation. 
Another method : 

V 




The equation to AB is - 7 ^- r -+• -4=- = 1 
OA OB 



Now 
Also 



OD = OAcosa.\OA = 



P 



COS a 



OD = OB cos DOB 

= OB sin a, .'. OB = 



sin a 



.-. the equation becomes 



+ 



y 



or x cos a -f y sin a 



(JL\ [_*_) 

\COS a/ \sin a/ 
j9, as before. 



= 1, 



EXERCISES. 

1. Find the slope form of the line through (1, 2), (5, - 
Suggestion. — Let?/ = mx + b be required equation. 

Then, 2 = m • 1 + b ) . , , , 

7 J solve for m and 6. 
_9 = m . 5 + 6) 



54 ANALYTICAL GEOMETRY 

2. Find the symmetrical equation of the line through (4, — 5) and 

(-3,7). 

Suggestion. — Let - + \ = 1 be equation required. 

a b 



Then, 



4 _ 5 

a b 

a b 



- solve for a and b. 



3. Find the normal equation of the line through (2, — 5) and ( — 3, 4). 
Suggestion. — Let x cos a+ y sin a = p be equation required. 
Then 2 cos a — 5 sin a = p (1) 

and — 3 cos a + 4 sin a = p (2) 

cos 2 a + sin 2 a = 1 (3) 

Solve (1), (2) and (3) for cos a, sin a, and p. 

4. Find the equations of the lines joining : 

(1). (1, 2) and (3, 4). (2). (- 2, 3) and (3, - 4). (3). (4, - 6) and 
(_2,-5). (4). (a, b) and (6, a). (5). (a + 6, o) and (o, a-b). (6). (3, 4)and 
(5, 2). (7). (-2, - 5) and (6, - 3). (8). (a + 6, b + c) and (a - b, b - c). 
(9). (2, 4) and (3, 6). (10). (0, 0) and (a, 6). 

5. Construct the following lines, (1) x = ^ 3, y = J^ 5, rr^.v^ 0. 
(2) :r + T/ = 1, 3x-y + 4 = 0, 2x + 3i/-M =0. 

6. What are the equations of the lines through the origin, and in- 
clined respectively at 45°, 60°, and 120° to the x-axis ? 

7. Find the equations of the lines, (1) through the point (1, 4) and at 
45° to x-axis ; (2). through (— 2, 3) and at 30° to sc-axis ; (3). through 
(2, — 5) and at 60° to x-axis ; (4). through (3, 4) and at 120° to x-axis ; 
(5). through (4, — 6) and at 150° to sc-axis. 

8. Reduce the following equations, first to the slope form, then to the 
symmetrical form. 

(1) 2 x - 3 y + 4 = 0. (2) x + 3 y - 9 = 0. (3) y - x = 4. 

(4) 8 x + 6 y = 5. (5) x - 5 y = 13. 

9. Find equations of these lines : 

(1) through (1, 2) and parallel to y = 3 x + 4 ; 

(2) through (—5,-3) and parallel to x — y = ; 

(3) through (6, — 4) and parallel to 2 x — 5 y = 16. 
Suggestion. — In (1) let y = Sx + k be line required. 
Then, 2 = 3.1 + fc, .-. k == - 1. 

.'. y — Sx — 1 is equation required. 



THE STRAIGHT LINE 55 

10. Find the intercepts of : 

(1) hx + ky =\. (3)8x-2y = 4. 

(2)4x-5y=12. (4) y-3x.= 2. 

(5) 2 ax + 5 6y = 13. 

11. Find the line through (4, 6) which cuts off equal intercepts on 
the axes. 

12. Find the line through (3, 3) which forms with the axes a A whose 

area is 18. 

x II 
Suggestion. — Let - + ^ = 1 be line required. 

Then, ab = 36 .1 . . (1) 

3 3 I Solve for a and 6. 

26. Any equation of the first degree in two variables, 
x, y, represents a straight line. — Take the most general form 
of such an equation, viz. : 

Ax + By + C = 0. 

Let (x x , y x ), (x 2 , y 2 ), (x 3 , y s ) be any three points on the 
locus represented by this equation, whatever it may be. Then, 

Ax, + By x 4- C = (1) 

Ax 2 + By. 2 + C = (2) 

A* 3 + By 8 4- C = (3) 

We now propose to eliminate A, B, and C from these three 
equations, and, if possible, to find some condition among the 
co-ordinates of the three points assumed, which may be in- 
terpreted in terms of known facts. 

Subtract (2) from (1), then (3) from (2), obtaining 

A(^-x 2 )+B(y l -y 2 )=0 ... (4) 
A(^-a0+B'(y,-y^=0 . . . (5) 

Eliminating B from (4) and (5), and dividing the resulting 
equation by A, we get 

(*1 - X 2) (ft ~ 2/3) ~ fa ~ *«) (j/l -2/2) = . (6) 



56 



or, 



ANALYTICAL GEOMETRY 




x x y x 1 






* 2 y% i 


= o, 




x s y z 1 





which is the condition that the three points lie on a straight 
line. Hence, any three points on the locus of Ax -f- By + C 
= are collinear ; i.e., the equation represents a straight 
line. Q.E.D. 

27. The theorem of the preceding article may also be 
proved indirectly, and somewhat less rigorously, by show- 
ing that the equation Ax -f- By + C = can be reduced to any 
one of the type forms which we have found in this chapter. 



Thus, I. 



By = — Ax 
A 



C 2 . 

c 

B 



Or, 
Whence 



II. If 



III. If 



IV. 



(which has the form y = mx + b). 
B = 0, then Ax + C = 0, 
x = — — (torm x = a). 

G 



Ax 



A 
By 



C + C 



+ 



y 



= 0,2/ = 

= -1, 

= 1 



(form y = b). 



f symmetrical form 



28. Reduction of Ax + By + C 



# ?/ 
a b 



to the normal form. 



If 

and 



Ax + By + C = 
iC COS a -f 2/ sin a — jo = 



(1) 

(2) 



represent the. same line, they must differ by some constant 



_\ _ C 2 



THE STRAIGHT LINE 57 

factor ; i.e., the coefficients in both equations must be pro- 
portional ; 

A _B_ _C 

COS a Sin a p 

A 2 B 2 _ / A 2 + B 2 \ C 2 

cos 2 a sin 2 a \sin 2 a -f- cos 5 

.•.A'+S- = | .... (1) 

J^ = 2! (2) 

sin 2 a p 2 

A 2 C 2 
-V = - 2 (3) 

COS 2 a p l 

C 
From (1), p = — , • 

V ; VA 2 + B 2 

From (2), sin a = — = — — = . 

C VA 2 + B 2 

From (3), cos a = • • 

V J VA 2 + B 2 

Hence the normal form of Ace -|- By + C = is 

A B C 

VA 2 + B 2 X VA 2 -4- W V " VA 2 -f B 2 ' 

Hence, to reduce the above equation to the normal form, 
divide through by the square root of the sum of the squares 
of the coefficients of x and y. It is convenient to change 
signs in Ax + By + C = (if necessary) so that C may be 
negative ; i.e., the right member of the normal form may be 
positive. 

Note 1. — Observe that the fractional coefficients of # and y above 
may be assumed to be the cosine and sine respectively of some angle, 
since the sum of their squares is equal to unity. 



58 



ANALYTICAL GEOMETRY 



Note 2. — The generality of the proof of the normal form of the 
equation depends on the fact that whatever the magnitude of a, OA and 
cos a have the same sign ; and OB and sin a have always the same sign. 
It is assumed generally that a is a positive angle. 





Fig. 66. Fig. 67. 

Note 3. — The symmetrical equation may be derived as follows : 
2 A AOB = 2 A POB + 2 A PO A. 

.•. ab = bx + ay, 



a b 

x %i 
29. Reduction of y = mx + b, and — \- - 

a o 

normal form. — These evidently become, 

m 1 b 

x+ t^ - = y — 



Vl + 



and 



bx 



+ 



VI + m 2 
ay 



Vl -f- *m 2 
ab 



1, to the 



0. (1) 



= . (2) 



Vtf 2 + b 2 ' \la 2 + b 2 -\fa 2 + b 2 

30. Digression on Algebra. — If we are given two equa- 
tions, as 

Ax + By + Cz = . . . . . (1) 

and A x x + B t y 4- C x z = (2) 

it is proved in books on Algebra that 

x y z 



B C 




C A 




A B 


B.C, 




0, A, 




A, B t 



THE STRAIGHT LINE 



59 



or, more simply, if z = 1, 



X 


y 




1 


B C 1 


C A 




A B 
A, B x 



0.) 



Hence, this last equation determines the point of inter- 
section of the lines 

A#'+ By + C = 0, 

Aja: + B lV + C 4 
We shall make use of this fact presently. 

31. Condition for the concurrency of three given lines. — 

Let the given lines be, 

Ax + By + C = (1) 

A t x + B lV + C, = (2) 

A 2 x + B 2 ?/ + C 2 = (3) 

Solving for the point of intersection of (2) and (3), we get, 

y 



B, C, 




0, A, 




K b, 


= A (say). 


B 2 C 2 




C 2 A 2 




A 2 B 2 




w x = X 


B s 


c 2 


. y = 


= X 


( 
( 


3, ^ 

-^2 ^2 


> 


L -X 


A X B X 

A 2 B 2 



If the three lines are to be concurrent, this point must lie 
on (1). Substitute these values in (1), and divide by X, 



or, 



= 0, 



B, C, 

B 2 C 2 


+ B 


c 2 


A 1 +C 

A 2 1 


A, B, 

A 2 B 2 




ABC 








A. B, C, 


= 0, 






A 2 B 2 


c 2 







which is the condition that the three lines given may be 
concurrent. 



60 



ANALYTICAL GEOMETRY 



Note. — The line y — y l = m (x — xj may represent any line of a 
pencil through the fixed point (x v y x ). 

Example. — If it passes through the point ( — 3, 4), we have 
y — 4 = m (x + 3) ; 
and if it makes 45° with rr-axis, we have 

y _ 4 = 1 (x + 3), 
or, y — x — 7 = 0. 

32. Angle between Two Lines, y = mx + b, and y = m'x 

+ b'. 

4> = e - e f . 




tan <£ — 



m — m 



Fig. 68. 

and 

they are parallel if 

or, 



1 -4- mm' 
If the lines be \\, cf> = 0, 
.'. m — m r = 0, m = m r . 
If the lines be _L, tan <f> = oo, 

.'. 1 -r- mm' = 0. 
If the lines are given in the 
form 

Ax + By + C = 0, 
A'x+ B'y+ C'=0, 

A A' 

B~W 



A B 
A! B' 



= 0. 



AA' 



They are _L if 1 + mm! — 0, or 1 + ^=- t = 0, 

-DO 

or AA r + BB X = 0. 

33. To Find the Equation of the Line through the Point 

(x v yj and _L to the Line Ax -f- By + C = 0. 

Let y — y t = m' (x — x x ) be the required equation. 



THE STRAIGHT LINE 61 



Then m — ——,.'. m' = I ) 

1> \ mj 



B 
A 



V-Vx =^( x - x i)> 






is the required equation. 



EXERCISES. 

1. Reduce the following equations to the normal form : 

(1) x + y = 3. (2) 5 x - 12 y + 20 = 0. 

2. Show that the lines, — h ~ = 1* ■=- + — = 1, and x = y are 

a o o a 

concurrent. 

3. Find the area of the A whose sides are 

x = y, x=—3y, x + y = — 4. 4ns. 8. 

4. Find the equation of the line joining (a, 6), and (—a, — 6). 

5. Find the equations of the medians of the A whose vertices are 
(2,1), (3, -2), (-4, -1). 

6. Show that the three points (1, — 1), (2, 1), and (—3, — 9) are 
sollinear. 

7. Find the equation of the line through (3, 4) and at 75° to the 
x-axis. Arts, y — 4 = (2 +^3) (x — 3). 

8. Reduce the following equations to the symmetrical form : 

(1) 3 x + 2 y + 6 = 0. (3) y - 4 x = 6. 

(2) 3 x - 2 y + 12 = 0. (4) - y - * = 2. 

(5) 6x + 5y= 13. 

9. Reduce to the normal form : 

(1) 3x + 4?y = 12. (3) 4x- y = 0. 

(2) x-^2y= 3. (4) x-2z/ = 1. 

(5) a; + y = 2. 

10. A straight line passes through the point (3, 4) and is bisected at 
that point. Find its equation. 

11. What system of lines are represented by the equation x cos a + 
y sin a = 9 if a be varied ? 



62 



ANALYTICAL GEOMETRY 



12. By the equation y = rax + 2, if m be varied ? 

13. Reduce to the normal form : 
£ , V 



+ r = i» 



y = mx + c. 



5 + »-i, 

a a 



t = 1. .4 ns. tan = =-; = 

a o 6 Z — cr 



14. Show that the lines 2 y — 3 x = 7, and 3 y + 2 x = 11, are ± to 
each other. 

15. Find the angles between the following pairs of lines. 

(1) 3 x - 4 y = 2, 2 x - ?/ = 0. ^.ns. tan <p = — |. 

(2) 

(3) x - 2 ?/ = 0, 2 x + 3 y = 1. 

16. Find the line through the point (— 1, — 1) and ± to the line y — 
2x = 1. 

17. Find the foot of the ± from origin to 3 x — 5 y = 4l. 

18. Find the line from (2, 3) and ± to x — y = 0. 

19. Find the line through (4, — 6) and parallel to x = 2 y. 

34- To Find the Distance from the Point (x v y x ) to the 
Line x cos a + y sin a = p. 

Y 



X 


A 




B 


V^A) ^ 









^S^ 1^ 



&i,yi)P 



Fig. 69. Fig. 70. 

AB is the given line, P (x v y x ) the given point, and A'B' a 
line through P parallel to AB. 

The equation to A'B' is evidently x x cos a -f- y x sin a = p t . 
In Fig. (1), 

PM = DD' = p x — p = x x cos a -\- y x sin a — p. 
In Fig. (2), 

PM = D'D = p — p x — p — x x cos a + t/j sin a. 



THE STRAIGHT LINE 



63 



Hence, the distance from the point P (x lf y x ) to the line 
x cos a -f- y sin a = p 
is equal to ± \x x cos a + y x sin a — jj], 

the upper or lower sign being used according as the point 
and the origin are on opposite sides of the line or on the 
same side. Notice that the required expression is found by 
writing the equation of the line so that its dexter is zero, and 
then substituting the co-ordinates of the given point for x 
and y. 

Again, if the equation be given in the form 

Ax -f By + C = 0, 
the distance is evidently 

d = Ax, + By t + C 
VA 2 + B 2 

35. Equations of Bisectors of the Angles between the Lines, 



Ax + By + C = 
A x x + B,?/ + Cj = 

/ 



(2) 




Fig. 71. 

If ~P (x lt 2/,) be any point on either bisector, it is equi- 
distant from the sides of the angle, 



64 



ANALYTICAL GEOMETRY 



. As, + By , + C = A^ + B^ + C^ 
VA 2 + B 2 "'" VA^+B, 2 

in which the upper sign is used for the bisector of the angle 
in which the origin lies. 

If the lines are given in the form 

x cos a -\-y sin a =p ...... (1) 

x cos aj -\-y sin a t = p x (2) 

the bisectors are 

x x cos a -\- y l sin a — p = db (^ r cos a t + i/j sin a r — ^J. 
Note. — The subscripts to the letters x, y, may be dropped in the end. 

36. Equation of a Line Through a Given Point (x xi y t ) 
and Making a Given Angle <£ with the Line y= mx + b. — 
There are obviously two lines fulfilling the conditions. AB 
is the given line. 

B 




Fig. 72. 

Let the required equation be 

y — y 1 = m' (x - ajj). 
Then, <f> = 0- r ,ov Q" - 0, 



tan <£ = 



???■ 



m 



m 



m 



-> or 



.*. m 



1 -f mm' ' f- mm' 

, m + tan <f> 
1 =f m tan </> 



THE STRAIGHT LINE 



65 



.'. The required equation is 

m ± tan <j> 
y ~ y^~ 1 qp m tan <£ 



(03 - 03,). 



37. Oblique Axes. Equation of straight line in terms of 
intercept 5, and angle 6, 
which the line makes 
with the z-axis. 

PD sin 6 



BD sin (<o — 0) 



or, 



y — 5 



sin 6 



x sin (o) — 0) 

sin 




If we put 
When 



Fig. 73. 

= m, we have y = mx -+- b. 



sin (w — 0) 

<o = 90°, m = tan 0. 

38. Equation of line in terms of one point (x v y A ) and 0. 

T 




PTx.i/) 



Fig. 74. 

Let P (aj, y) be any point on the line. 

sin 



Then 



or, 



PD 



CD sin (o> - 0) 

y- y t _ sin g 
# — a^ sin (<y a 0) 



ANALYTICAL GEOMETRY 



39. Equation of a line in terms of the _L on it from the 
origin and the angles which this _L makes with the axes. 

The equation of AB is 




OA ^ OB 



But OA = 



P 



COS a 

OB= * 

cos /? 

where p = OP. 
.*. x cos a + y cos /? = p 
is the required equation. 

40. Polar Co-ordinates. Equation of the straight line 
through two points (p v 6 X ) and (p 2 , 6 2 ). 




or, 



Fig. 76. 

A OBC = A OBP + A OCP. 

•"• i Pi P2 sin (0 2 ~B X ) =\ PPx sin (0 - t ) 

+ I P Pa siri (A - #)• 
.'. PPl sin (0 — t ) + PlP2 sin (0 t - 2 ) + /o 2 /o sin (<9 2 - 0) = 0, 
sin (fl t - fl 2 ) , sin (fl 2 - 0) sin (6> - y ) = Q 

P />! P% 



THE STRAIGHT LINE 



67 



This result can be easily remembered in the determinant 
form, viz. : 



P 
1 

Pi 
1 

P-2 



COS 



cos 1 sin t 



cos 6o sin O 



0. 



41. Equation in terms of the _L on the line, from the pole 
and the angle which it makes with the initial line. 

The J_ OD = p, 
Z DOC = a, 
OA is the initial line. 

J? = cos POD, 

P 

or J - = (cos 6 — a). 

n v J Fig. 77. 




r cos (0 - a) ' - 5 
which is the equation required. 

Discussion-. (1) If the given line CB is _L to the initial line a = 0, 
and the equation becomes 

_ -P 
P ~ cos ' 

which is the equation of a line perpendicular to the initial line. 



* Note. — This equation 

p = p cos (d — a) 
gives p = p [cos 6 cos a + sin 6 sin a] 

= (p cos 6) - cos a + (/> sin d) • sin o, 
.* . p = x cos a + y sin a, 
which agrees with § 25.' 



68 ANALYTICAL GEOMETRY 

(2) When, = 0, p - ^f-^ = 33^ _ 0C . 

(3) When 0= a, -P^-P- 

IT X> 

(4) When 6 = - + a, p = ^ = oo , for in this case the radius 

vector becomes parallel to the line. 

(5) When 6 = 2 it + a, P = P (as it should be). 

(6) When = 2 *■, /o = ^ = OC. 

v ' cos ( — a) 

(7) When AB passes through the pole 0, p = 0, which is for every 
value of except - + a, when /o assumes the indeterminate form ^- • 



EXERCISES. 

1. Find a line making an angle of 60° with the line 5z + 12?/ + l = 0. 

, 12^3 + 5 

Ans. y = — x + b. 

5\/3-12 

2. Making an angle, tan- 1 - with the line - + ^ = 1, and passing 

through the point ^- a, - bj • Ans. y-^ = yTZtf (* ~~ l) * 

3. Show that the A (2, 1), (3, - 2), (- 4, - 1) is a right A. Prove 
analytically that, 

4. The altitudes of a A are concurrent. 

5. Th'. medians of a A are concurrent. 

6. The ± bisectors of the sides are concurrent. 

7. The bisectors of the angles are concurrent. 

8. The orthocenter, centroid, and circum-center are collinear. 

9. Find the locus of a point whose distances from the lines 

A x x + B t y + C, =0, A 2 x + B 2 y + C 2 = 0, 
are in the ratio m : n. 

. A,x + B,y + C l . A 2 x + B 2 y+C 2 

Ans. The line n— — , — I = ±m — - — 2 * — ?• 

VA, 2 + B x 2 VA 2 2 + B 2 2 

10. Show that the angle between the lines x — y\/S + 1 = and 
% + yVS - 2 = is 60°. 



THE STRAIGHT LINE 69 

11. Show that the lines 6 x + 4 y = and 2x — 3y -1 = are ± to 
each other. 

12. Show that Ax + By + C : =0, and Bx — Ay -f- C 2 = 0, are ± to each 
other. 

13. Show the same of the lines ax + by + c = 0, and 

x y 

14. Find the ^/s between these pairs of lines. 

(l)5y-3aj = 0, y-4x = l. Ans. 45°. 

(2) 2x-3y = 0, 6x + 4y + l=0. ^Ins. 90°. 

(3) y + £ = 0, (2 + V / 3)y-a; = 0. ^4ns. 60°. 

(4) 2/ = A-ar, (l-k)y -(1 + k) x = 0. Ans. 45°. 

15. Find the line through (2, 3) and ± to the line joining the points 
(1, 2) and (-3, - 14). 

16. Find the line through (0, — 1) and X to X + y = 1. 

-4ns. x — y = 1. 

17. Find the line through the origin and ± to the line Ax + By -f C = 0. 

Ans. Bx = Ay. 

18. Find the distance of the line h (x + h) + k (y + k) = from the 
origin. AnSm x / ~W~~+~kT 

19. Also, the distance of - + - = 1 from origin. Ans. 



Ans. V226. 



h k V h 2 + k? 

20. Show that .the origin is inside the A whose vertices are (3, 4), 
(2, -3), (-2, -2£). 

21. Find the distance between the feet of the ±s from the origin on 
the lines, 

3x-4y + 25 = 0,) 
12x + by - 169 = 0. J 

REVIEW EXERCISES. 

1. Which quadrants do the following lines cut ? 

(1) y=2x+ 3. (3) y= - x - 5. 

(2)?/= -5x + 8. (4)y= 2x-3. 

(5) x- by - 9 = 0. 

2. Reduce to the normal form the equation 

3x + 4y - 15 = 0. 



70 ANALYTICAL GEOMETRY 

3. Axes make an /_ 60°. Find a line, intercept on y-axis — 4, and at 
30° to x-axis. Ana. y — z — 4 = 0. 

4. Find the line through (0, 0) and (3, 2) . 

5. Find the intersection of lines 

3?/ + 4x-ll = 0, 
4 y + 3 x - 10 = 0. 

6. Find locus of a point equidistant from (1, 1) and (— 1, — 1). 

Ans. x + y = 0. 

7. Find a line equidistant from x + 1 = and x = 3. ^4ns. »= 1. 

8. Also equidistant from y = k and y = c. Ans. y = § (c -f &). 

_£ ^ 9. EF and GH are parallel to 

the sides of the £? . Find the locus 
of the intersection of GE and FH. 

Ans. A diagonal. 




Suggestion. — Take AB and AD 
as axes. 



F B 

Fig. 78. 

10. Find the bisectors of the angles between the lines 

3 x - 4 y + 7 = 0, ) Ans. 21 x + 77 y = 116, 
12 x + by - 5 = 0. j 33x-9y=-22. 

11. Show that one of the bisectors of the angles between the lines 

3 x — 4 y = 5, 
5x +12 y =13 
passes through the origin. 

12. Find the bisectors of the angles between the lines 



u 



4x+ 3y = 3, ) -4ns. 77 x - 21 y = 79, 
5x-12y = 8. J 27x + 99y+l = 0. 

42. To determine the relative positions of two given 
points (x lf y t ) and (x 2 , y 2 ) with respect to the given line, 

Ax + B?/ + C = 0. — Let us suppose that the line divides 
the distance between the points in the ratio of m : n. Then 
the co-ordinates of the point of division are, 

mxo + nx x miio + wvi 

x = ; > y = , — — • 

m -+- w. m + w 



THE STRAIGHT LINE 71 

These must satisfy the equation Ax + By 4- C = 0. 
Substituting these values and multiplying by (m + n), 

A (mx 2 + nx t ) + B (my 2 4- ny t ) 4- C (m 4- n) = 0, 
or, m (Ax 2 4- By 2 + C) = — n (Ax x 4- B < j/ 1 4- C). 

. m _ _ As, + B?/ x + C _ 

' ?i ~ A# 2 -f- B?/ 2 -f- C 

Now, if (x l , y^, (ai. 2 , y 2 ) are on the same side of the line, 

itl 
the distance between them is cut externally, .'.— is negative ; 

.*. Ax x 4- By i + C, and A.r 2 + By 2 -\- C, have like signs. If 

they are on opposite sides, — is positive, and .*. Ax x + By^ 

4-C, and Ax 2 4- By 2 + C, have unlike signs. Hence two points 
(a^ , yj, (ic 2 , ?/ 2 ) are on the same or opposite sides of Ax 4- 
By 4- C = according as the results of substituting their co- 
ordinates in the sinister of the equation have like or unlike 
signs. 

Note. — Observe that equation (1) gives the ratio into which the 
distance between the points is divided by the given line Ax + By + C = . 

EXERCISES. 

1. Find the area of the A formed by the lines 

V = Az x i x — c > Ans. c?. 

2. Show that the angle between the lines 

x + yVS = 0, j 
x - y V3 = 0, ) 

3. Find the line through (4, 5) and parallel to 2x — 3 y = 5. 

4. Show that the distance between the points (1, 2), (4, 3), is bisected 
by the line joining (2, 3) and (4, 1). 

5. Find the distance of the line -+ | = 1 from origin. 

2 o 



72 ANALYTICAL GEOMETRY 

6. Find the polar co-ordinates of the intersection of the lines 

2a 
P = 



(cos 6 - I) 



(-!)] 



Ans. p = 2 a, 




Show that they meet at an angle -• 

o 

7. Show that the area of the A whose sides are 

x+2?/-5 = 0, 
2x + y -1 
y — x — 1 : 

8. Show that the line 4x-f 5y = 6 cuts the distance between 
(3, — 2), (1, 2) in a point of trisection. 

9. Show that the points (1, — 2) and (0, 1) are on one side of the liue 
in Ex. 8. Also show that the origin and the point (1, 1) are on opposite 
sides of the line. 

10. Show that the origin lies inside the A whose vertices are the 
points of intersection of the lines by — 4x — 1 = 0, x — Sy — 9 = 0, 
x + 9 y — 9 = 0. Prove the same of the point Q, ^). 

11. Find the bisectors of the angles between the lines 

4y + Sx-9 = 0,) Ans (99x+27y-87 = 0, 
12x- 5y + 6 = 0.) \ Sx- Uy + 21 = 0. 

12. Show that the distance from the origin to the line 

= 1 igl 

p 3 cos 6 + 4 sin d 3 ' 

Also, its distance from the point ( 13, tan-i _j is 11. 

Show also that the equations of these J_s are 

4 „ 33 

6 = tan— i ^ , and p = 6 , , . — = — r- 

3 r 3 cos 6 + 4 sin 

13. Find the angle between the lines 3x + y+12 = and x + 2 y — 
1 = 0. 

14. Find the distance from (4, 5) to the line 4 x + 5 y = 2„ 

15. Find the distance from (1, 2) to ^ - 5 x = 18. 



THE STRAIGHT LINE 73 

16. If the axes make an angle of 60°, find the angle between the line 
y — 2 x — 5 = and the x-axis. Ans. tan£\/3. 

17. If the axes make an angle of 45°, find the angle between the lines 

7x + 3y-l=0, x + ?/ + 2 = 0. 2 V2 

Ans. tair 



5 (V2 - 2) 

18. Show that the lines x + y = m, x — y = n, are JL to each other 
whatever the angle between the axes. 

19. Show that the lines in Ex. 17 make these angles with the x-axis, 

7 



tan-i 



tan - 1 



7- 3 V2 
1 



1 - v x 2 

20. Find the locus of the vertex of a A having a given base and a 
given area. 

21. Find the locus of a point equidistant from the points (x t , y{) and 

(«2, 2/2). 

43. Equations representing straight lines. — An equation 
whose dexter is zero, and whose sinister can be broken up 
into factors of the first degree, represents straight lines. 

Examples. — 1. 2x 2 +xy — 3?/ 2 = represents two lines, for it may- 
be written, (2x + 3 y) (x — y) = 0. 
.*. the lines are 2x + 3y = 0, x — y = 0. 

It is evident that the co-ordinates of all points on the two lines satisfy 
the given equation. 

2. 2x 3 + ax 2 — 7 a 2 x — 6 a? = represents three straight lines parallel 
to the y-axis, for it may be written (x + a) (x — 2 a) (2 x + 3 a) =0. 

3. 2xy+4x — Sy — 6 = represents two lines, for it may be written 
+ 2) (2x-3) =0. 

4. What is represented by the equation 

(Sx + 2y-l) (2X + 2/-3) =0? 

44. A homogeneous equation of the nth degree represents 
n straight lines through the origin. — Take the equation, 
Ax n 4- Bx"- 1 y 4- Cx n ~ 2 if + • • • • 4- l&y n = 0. 



74 ANALYTICAL GEOMETRY 

Dividing by Ay n , we get 
fx\ n B /x\ "- 1 C fx\ n ~ 2 N 

Now if m x , ra 2 , ra 8 • • • ■ 7?i n denote the roots of this equa- 
tion, then 

g-™ 1 )g-- 2 )g--e)--g-^)=0. 

Hence this represents n straight lines through the origin ; 
viz., x— m x y = 0, x — m 2 y = 0, x — m n y = 0, etc. 

Example. — Show that the equation 
A<y-y i T+B(y-y l ) n -\x-x 1 ) + C(y-y i f-\x-x 1 ) 2 +. • -+^(x-x 1 ) n = 
represents n straight lines through the point (x v yj. 

45. The homogeneous equation of the second degree, — 

This represents two straight lines through the origin. 
Take Ax 2 + 2 Hxy + By* = 0. 

y 

Divide by x 2 , giving a quadratic in-, viz., 
■i[^Y+2H f^) + A = 



x \x 



If m 1 and m 2 denote its roots, we have 

- H + VH 2 - AB - H - VH 2 - AB 
m 1 = g , m 2 = g 

Hence the equation represents the two lines 
y = m^, y = m 2 x. 

46. Angle between the lines Ax 2 + 2 Hry + By 2 = 0. 

m, — m 



tan <£ = 



1 4- mj' 


w 2 


2VH 2 - 


- AB 


B 


2 VH 2 - 


- AB 



4**] 



A + B 



THE STRAIGHT LINE 75 

Discussion. (1) If the lines are _L, then A + B = 0. 

(2) If the lines are ||, then H 2 - AB = 0, 

which makes the two lines coincident. This is also obvious since both 
lines pass through the origin. 

Note. — A pair of lines through the origin at right angles to each 
other may be represented by x 2 + X xy — y 1 = 0, where X is a variable 
parameter. 

47. The bisectors of the angles between the lines Ax 2 + 
2 H.xij + Bif = 0. 

Write the equations of the lines thus : 

V = ™ x x (!) 

V = m 2 x (2) 

Then the equations of the angle bisectors are 
y — m x x y — m 2 x 



Vl + m 2 Vl + 



lllr, 



.... (3) 



y _ miX y -m 2 x = 



Vl + m 2 Vl + m 2 * 

These two bisectors may be represented by a single equa- 
tion, whose dexter is zero, and sinister is the product of the 
sinisters of these equations, viz., 

(y - ™ y xf (?/ - tn 2 xf 







1 + m t 2 


1 + m 2 






vyj 


or 


(m 


2 -m 2 )(y 2 — x- 


; ) 4- 2(m 1 m 2 — 1) (m, — m 2 )xy = 


= 


(6) 


or 




(m 2 + ™>^(f 


- x 2 ) - 


■ 2 (m 1 m 2 


- 1) xy = 




(') 






• 2H ^ 2 


-X 2 )- 

x 2 - 
' ' A - 


•<- 

■ y 2 xy 

- B H 


l)xy = . 




(8) 
(9) 



Equation (9) represents the two bisectors. 



76 ANALYTICAL GEOMETRY 

48. To Find the condition that the general equation of 
the second degree, may represent a pair of straight lines. — 

Let the equation be written, 

Ax 2 + 2 Hxi/ + By 2 + 2 Gx + 2 Fy + C = . (1) 

.\ Ax 2 + 2 (Hy + G) a; + By 2 + 2 Fy + C = 0. 

Solve for x, obtaining, 



Ax + Hy + G = ± V(Hy + G) 2 - A(By 2 + 2 Fy + C) 



= ± V(H 2 -AB)y 2 +2(GH-AF)y + G 2 -AC. 
The expression under the radical is a perfect square if 
(H 2 - AB) (G 2 - AC) = (GH - AF) 2 , 
or, ABC + 2 FGH - AF 2 - BG 2 - CH 2 = 0. 

This expression is called the discriminant of (1), and is 
usually denoted by the symbol A (delta). It may be easily 
remembered in the determinant form, viz., 

A = A H G =0. 



A 


H 


G 


H 


B 


F 


G 


F 


C 



Hence this is the condition that (1) the general equation of 
the second degree may represent a pair of straight lines. 

Example. — Determine k so that the equation 

x 2 + kxy + 5y 2 - 6 y + 4x + 5 =0 
may represent two straight lines. 

EXERCISES. 

1. Show that the line joining the points (3 cos a, 2 a) and (3 cos 2 a, 
3 a) is p = 3 COS (0 — a) . 

2. Show that the line from (p., 0.) ± to p = ; r is, 

yHv v H cos (0 - a) ' 

p sin (a — 0) = p 1 sin (a — : ) . 

3. Show that y + x = and y — x = are always ±. 



THE STRAIGHT LINE 77 

4. If the axes make an angle of 30°, find equation of the line making 
105° with x-axis. Ans. x + y — 1 = 0. 

5. Find the angle between the lines 
7y - 17 x - 1 = 0. 



Ans. cos- 1 1|. 
x - 2 = 0. l 

6. Find the bisectors of the angles between the lines 

3x 2 + 8xy + Sy 2 = 0. ^ns. x 2 - y 2 = 0. 

Also, between the lines 3 x 2 + 4 x?/ — 5 y' 1 — 0. Arcs, x 2 — ixy — y 2 = 0. 

7. Show that the lines joining the origin to the points of intersec- 
tion of y = k (x — 4), and ?/ 2 = 4 x, are ±. 

Suggestion. — The lines are k (x ? — y 2 ) = xy. 

8. Show that 3 x 2 + 2 xy — 3 y 2 = are two _L lines, 
and x 2 + 2 xy + Z/ 2 — £ — y — 6 = are two || lines. 

9. Determine k so that Qxy — 2x + ky + 5 = may be two straight 
lines. Ans. k = — 15. 

10. Show that the lines Ax 2 + 2 Hx# ■+■ By 2 = k (x 2 + y 2 ) are equally 

inclined to the lines 

Ax 2 + 2 Hxy + By 2 = 0. 

Note. — Observe that they have the same angle-bisectors. 

49. Lines through the point of intersection of two given 
lines. — If S x == and S 2 = 0, represent the equations of any 
two loci, then the locus of S x + kS„ = passes through all 
the points of intersection of the two given loci ; for the co-ordi- 
nates of any points common to both evidently satisfy the last 
equation. 

This principle may be applied to the straight line as 
follows : 

Example. — 1. Find the equation of the straight line joining the 
point (1, 2) to the intersection of the lines 

x-2y + 6 = (1) 

and 2 x -f y = (2) 

Let the required equation be 

(x - 2 y + 6) + k (2 x + y) = (3) 

which evidently may represent any line of a pencil passing through the 
point of intersection of (1) and (2). We may find the required line by 
determining for what value of k (3) will pass through the given point (1,2). 



78 ANALYTICAL GEOMETRY 

Substituting its co-ordinates in (3), we get 

l-4 + 6 + 2fc + 2fc = 0, 
whence, * k = — f . 

.'. (x - 2y + 6) + (- |) (2x + 20 = 
is the equation required. It passes through the point (1, 2) and through 
the intersection of the lines (1) and (2). 

Note 1. — If we desire the equation of the line passing through the 

point of intersection of the lines 

Sx-0 (1) 

S 2 = (2) 

and parallel or perpendicular to a third given line 

S,= 0. ........ . (3) 

we assume the required equation as before, viz., 

S x + A;S 2 = (4) 

We find the slope of (4) in terms of k, and then ascertain for what value 

of k the line (4) will be perpendicular or parallel (as the case may be) to 

the line (3). Equation (4) (with the value for k substituted) will then 

be the required equation. 

Note 2. — The angle bisectors of the lines 

Sj = 0, S 2 = 0, are S, ± S 2 = 0, or S t =f S 2 = 0. 

50. To find the equations of two lines drawn from the 
origin to the points of intersection of the line 

with the locus of the general equation of the second degree, 
Ax 2 + 2 Hxy + By 2 -f- 2 Gx+ 2 Yy + C = . . (2) 
Consider the equation, 

Ax 2 + 2 Kxy + B y 2 -f 2 (Gx -f ~$y) 

(M)+ C (MJ=° • • • • (*> 

It evidently passes through the points of intersection of the 
given loci, for values which satisfy (1) and (2) satisfy (3). 

Again, (3) represents a pair of straight lines through the 
origin, since it is homogeneous and of the second degree. 



THE STRAIGHT LINE 



79 



Hence equation (3) represents a pair of straight lines joining 
the origin to the points of intersection of (1) and (2). 

51. The line at infinity. The intercepts of the line 

C C 



Ax + By -f- C = are 



B 



respectively. 



Discussion (1). — Now suppose A becomes very small; then 

C 



be- 



When A = 0, — = co , and the line 

.A. 



comes numerically very large 

becomes parallel to the x-axis. 

(2) If both A and B become very small, the line has very large 
intercepts. 

(3) If A = 0, and B = 0, both intercepts become infinite, and the 
line is situated altogether at infinity. Its equation then becomes C = 0, 
or k (a constant) = 0, which is the usual form of the equation of the line 
at infinity. 

52. Parallel lines meet at infinity. Let any two lines be 
given, as, 

Ax +B// +C =--0 (1) 

A^ + B^ + C^O (2) 

Then the equation of the line at infinity is, 

o . x + o • y + C 2 = (3) 

The condition that these three lines be concurrent is 



= 0, 



but this is evidently also the condition that (1) and (2) be 
parallel. Hence parallels meet at infinity. 

EXERCISES. 

1. Find the line through (a, b) and the intersection of 



A 


B 


c 


= 0, 






A, 


*i 


o, 




A 


B 








c 2 


or 


K 


B t 



XT/.-) 










5+f-M 


A X 

Ans. — - 


V 


1 


1 


x y -. f 


a 2 


b l 


a 


b 


b + a ' J 











80 ANALYTICAL GEOMETRY 

2. Find the line through the point of intersection of 

Ans. 88 y -66 x = 101. 

3. Show that the angle between the lines 

x 2 — 2 xy sec 6 + y 2 = is 6. 
Also, the angle between the lines 

6x 2 —xy — y 1 = is 45°. 

4. Find the line from (0, 3) to the point of intersection of 

y — x — 1 = 



y -2*-2 = o.' Ans - r-a«-a-o. 



5. Find the line joining the points (p v X ) and \2p v d x + J ) 



Ans. p 



cos (0 - 0J 

6. If the angle between the axes is \f/, show that the line through 
(h, k) and ± to the x-axis is 

x + y cos yjn == & + k cos ^. 

7. Find the distance from (6, a) on - + ^ = 1. 

8. If the axes are inclined at 60°, show that the distance from (3, — 4) 
to 4x + 2y = 7 is f. 

9. Show that the equation, sin 30 = 1, represents three lines through 
the origin and making angles of 30°, 150°, 270°, with x-axis. 

Note on imaginary loci. The locus of the equation x 2 + y 2 = is 
manifestly imaginary, for x ? and y 2 are positive, .-. their sum cannot 
equal zero. The only point satisfying the equation is x=0, y = 0. 
Hence this equation is sometimes used to represent the origin ; or it may 
stand for the "point-circle," i.e., a circle whose center is at the origin, 
and radius is 0. Also, since it may be written (x + y"s/ — 1) (x — yV ' — 1) 
= 0, it represents two imaginary straight lines. The equation 
(x-2/+l) 2 +(2x + 2/-2) 2 = 

represents a point found by treating these equations as simultaneous, viz., 
x - y + 1 - ........ (1) 

2x + y-2 = (2) 

or two straight lines (x — y + 1) = J- V 7 — 1 (2x + y — 2). 

The equation 9 (2 x — y — 5) 2 + 5 (x — y + 3) 2 = represents a point 
found as above, or two imaginary straight lines ; viz., 



THE STRAIGHT LINE 



81 



3(2x-y-5)+V-5(x-y + S) = . . . . (1) 
3(2x-y-5)-y-5(x-y + 3) = . . . . (2) 

Note on concurrent lines. If S x = 0, S 2 = 0, S 3 = 0, are the equations 
of three loci, and if we can find three quantities I, m, -and n, so that the 
sum ZS t + mS 2 + nS 3 vanishes identically, i.e., is equal to zero, then the 
three loci meet in a point ; for any points common to any two of these 
loci will lie on the third, since they will satisfy the equation 
ZS X + mS 2 + nS 3 = 0. 

This fact will sometimes be found very serviceable in the study of the 
straight line. 



EXERCISES ON CHAPTER IV* 

Show that the area of the A formed by the lines 

y = w, x -j- c,, "] 

y = m 2 x + c 2 , [ is \ 

y = m 3 x + c 3 ,\ 



(c 2 



^)_ 2 + ^ 



c ] Y + (c 1 



c 2 ) 2 



Obtain also the following expression for area : 



2 (ra t - m 2 ) (m 2 - m 3 ) (m 3 - mj 
2. Show that the area of the A formed by the lines 



0,1 



is \ 



B, 
B 2 

B, 






AjX + B x y + c x 

A 2 x + B 2 y + c 2 = 0, 

A 3 x + B 3 y + c 3 = 0, 

(A, B 2 - B 1 A 2 ) ( A 2 B 3 - B 2 A,) ( A 3 B x - A, B J . 

3. Show that the area of the A formed by the ?/-axis and the lines 

?y = m 1 x + c i ,) . (c 2 - c,) 2 
y = m 2 x + c 2 , ) 5 2(ra 2 — m^' 

4. Show that the area of the A whose sides are 
x - x x y + ax* = 0, "1 

x-a;,y + ax 2 2 = 0, L is £ a 2 j (x x — x 2 ) (x 2 — x 3 ) (x, — x,) j . 



x 



x 3 y + ax 3 2 = 0, 



* It is expected, that from the lists of exercises given in this book, 
the teacher will make his selections according to the needs of his class 
and the time at their disposal. 



82 



ANALYTICAL GEOMETRY 



5. Show that the area of the A whose sides are 
x 



-1 = 0. 

a b 

5 + S-p. 

a b 



sec 



tan 0=1, 



is ah. 



6. Show that the distances from the origin to the bisectors of the 
angles between the lines 

X cos a + y sin a = _p, 
x cos |8 + ysin /3 =p', 



are 



and 



/3) 2 cos i (a - /3) 

7. Show that the distance from the origin to the line joining the 
points (/o 2 , 2 ) and (p v t ) is 

P1P2' sin (fl 2 -*i) 
Vpi 8 + P2 a -2p 1 p 3 cos(tf 2 -tf 1 ) 

8. Show that the area of the parallelogram whose sides are 

A t x + B,y + C, + \ = 0, 
A 2 x + B 2 y + C 2 + k 2 = 0, 

k,k 



A,x + B t y + C x = 0, 

A 2 x+B 2 y + C 2 = 0, 



9. Show that the equation of the line joining the points (2 k cos a, a) 
and (2 k cos £, /3) is 

2 k cos a cos j8 

p = — • 

r cos (j8 + a - 0) 

10. Prove that the lines y = m 1 x + c v y — m 2 x + c 2 , are equally in- 
clined [in opposite directions] to the x-axis if 

1 = — 2 cos 0, 

m 1 m 2 

where <f> is /_ between axes. 

11. Show that the line joining the point of intersection of 

A,x + B,y + C t = 0,; 

A 2 x + B 2? / + C 2 = 0, 

A 3 x + B 3 y + C 3 = 0, 
A 4 x + B 4 y + C 4 = 0, 
A,x + B,y + Cj + & (A 2 x+B 2 ?/ + C 2 ) =0. 



to that of 



THE STRAIGHT LINE 



Where k is determined by the equation, 
A 2 B 2 
A 3 B 3 
A 4 B 4 

12. Show that the line from (p v d y ) and _L to the line 

1 



c, 




A, 


B, 


c, 


O n 


+ 


A, 


B 3 


c, 


c 4 




A 4 


B 4 


c 4 



IS 



ei 



a cos -f b sin 
6 cos d — asm 6 



cos X — a sm 0j 

13. Show that the angle between the lines 

3 x 2 + xy - 2 y 1 + x + 6 y - 4 = is tan -1 5. 

14. Show that the equation tan 3 0=1 represents three lines through 
the origin, viz., d = 15°, 75°, or 135°. 

15. Show that the condition that the line 

A x x + B,y + C, = 
may pass through the intersection of the lines 

Ax 2 + 2 Hxy + By 2 + 2 Gx 4- 2 Fy + C = 

A H G 
is H B F = 0. 

A, B, C, 

Note. — Observe that the points of intersection of the two given 
lines (if the given equation represents lines) are determined by any two 
of the equations 

Ax + Hy + G - 0, 

Hx + By + F = 0, 

Gx + Fy + C = 0. 

16. Show that the equation 

x 2 + 2 xy cos <f> + y 2 cos 20 = 
represents two straight Hues at 45°, and 135° to the x-axis [where <f> is 
the /_ between the axes]. 

17. Show that the pair of lines joining the origin to the points of 

intersection of 

X 2 + yt _ 2 x - 4y - 31 .= 0, 

and x + y —2 = 0, 

is 31 x 2 + 74 xy + 35 y 2 = 0. 



84 ANALYTICAL GEOMETRY 

18. Find the line joining the origin to the intersection of the lines 

AjX-f B x y + C, = 0, A 2 x 4- B 2 y 4- C 2 = 0. 

Ans (A, C 2 - A, d) x 4- (B x C 2 - B 2 Cj y = 0. 

19. The line - + ? = 1 

a b 

moves so that — \- =-= k [a constant]. 

Show that the line always passes through the fixed point [ j-", -r\ 

20. From P (h, k) ±s are drawn [PM and PN] to the axes inclined 
at an /_ <f>. 

Prove, MN = sin $ \Zh 2 4- k 2 - 2 hk cos 4> 

and the equation of the i. from P on MN is 

hx - ky - h 2 4- k 2 = 0. 

21. A line revolves about the origin and cuts the lines 

A x x + B, y 4- C t = 0, A 2 x 4- B 2 y 4- C 2 - 0, 
in P and Q respectively A point R fs taken in the revolving line such 
that OR = OP 4- OQ [O is origin] Show that the locus of R is 

jA,aj+ B& + C,jfA 2 x4- B 2 2/ + C 2 j=C 1 C r 

22. Show that the equation 

x 2 + kxy + y 1 - 5x - 7 y 4- 6 =0 
will represent two straight lines if k = f or ^. 

23. Find the line joining the origin to the point of intersection.of the 
lines 

a + b" lt 

o a 

24. Show that the lines Ax 2 4- 2 Hx?/ + By 2 = 
are ± to the lines Bx 2 - 2 Hxy 4- Ay 2 = 0. 

25. Show that the product of the J.s from (h, k) on the lines Ax* 4 
2 Kxy + By 2 ~ is 

Aft 2 '4- 2 HM 4- B& 3 



V(A - B) 2 4- 4 H 2 
26. Show that the equation 

(# 4. k 2 - 1) (x 2 4- y % - 1? = (Ax 4- ky - 1)» 
represents a pair of straight lines. 



THE STRAIGHT LINE '85 

27. Show that the equations 

A tan e + B sec d = 1 (1) 

and A sec 6 + B tan 6 = 1 (2) 

represent pairs of straight lines. 

Suggestion. — Transform to rectangular axes. 

Show, also, that the condition that the lines (1) should be the bisec- 
tors of the angles between (2) 
is A 2 - 2 B 2 + 1 = 0, 

and (1 - B 2 ) 2 + 2AB - 0. 

28. Find the points of intersection of the lines 



P= 7 =v' P = 



( a £\ 

\cosl5 ' 4/ 



And of the lines p = a sin 0, 



29. Prove that the distance between the points of intersection of the 
line x cos a + y sin a = p 

with the lines Ax 2 + 2 Hxy + By 2 = 0. 



2pVH 2 -AB 
is 



A sin 2 a — 2 11 sin a cos a + B cos 2 a 

30. Show that the lines 

A 2 x 2 + 2 H (A + B) xij + B 2 ?/ 2 = 
are equally inclined to the lines 

Ax 2 + 2 Hxy + By 2 = 0. 

31. The line to + my + n = meets the lines Ax 2 + 2 Hxy -f- By 2 = 
at angles a and £. 

, , (A - B) Im - H (Z 2 - m 2 ) 

Prove : tan a + tan 8 = 2 j-= — — ^— — - — - . 

H Al 2 + 2 Him + Bm 2 

Query. — Why does the quantity n not enter into the result? 



CHAPTER V 




THE CIRCLE 

53. Equation of the circle whose center is the point (h, k), 

and radius r. — 

P (x, y) is any point on the 

circle.* 

Then CP 2 = CB 2 + PB 2 
or, r 2 = (x - hf -f- (y - k)\ 

... (x - lif -f- (y-kf = r 2 

^ is the required equation of the 
ng. 79. circle. 

54. Special positions of center. — 

(1) If the center is at the origin, 

h — o, k = o, 
and the above equation becomes 

x? 4- J 2 = r\ 

T, Y 

Fig. 80. 

P(x,y) This is also evident from the 

figure given here. 
oo v) (2) When the circle passes 
through the origin and has its 
center on the #-axis to the right 
^ of the origin, r = h, k = o, and 
the equation becomes 

(x — r) 2 + y 2 = 7 s - 
or, x 2 -f y 2 a 2 re. 





F/ff. 81. 



* Circle, in this chapter, means circumference. 

86 



THE CIRCLE 



87 



(3) When the center is on the ^-axis above the origin, and 
the circle passes through the origin, we have h = .o, r = k. 

.-. the equation is x 2 + (y — r) 2 = r 2 

or, x 2 + 2/ 2 = 2 >■*/ 

Query. — What 0s are represented by the equations x 2 + y 2 = 
— 2 rx, x 2 + 2/ 2 = —2ry? 

55. Every equation of the form x 2 + !/ 2 + 2 Gx + 2 F?/ 
-j- C = represents a circle. — For this may be written thus : 

a* + 2 G x + G 2 + S/ 2 + 2 Fy + F 2 = G 2 + F 2 - C. 
.... (x + G) 2 + (y + *7 = G 2 + F 2 - C. 

Comparing this equation with (x — 7^) 2 + (y — &) 2 = r 2 , we 
see that it represents a circle whose center is the point 
(— G, — F) and whose radius is VG 2 + F 2 — C. Flence, 
an equation of the second degree in rectangular co-ordinates 
represents a circle if, 

(1) The term in xy is absent, and (2) the coefficients of x 2 
and y 2 are equal. 

Note. 1. We generally make the co- 
efficients of x l and y 1 unity by division 
if necessary. 

Note 2. The equation of the circle 
with center (r, o) and radius r may also 
be found as follows : 

PTJ 2 = OD x DA, 

or, y 1 = x (2r — x), 

whence x 2 + y 2 = 2 rx. Fig. 82. 




56. The conditions determining a circle. — Three condi- 
tions determine a circle, e.g., three given points through which 



88 



ANALYTICAL GEOMETRY 



it is to pass. Let the circle be x" -f y' 2 -f- 2 Gx -j- 2Fy -f C = 0, 
then the co-ordinates of each point, when substituted herein, 
give a relation between G, F, and C. Hence, from three re- 
lations (equations) G, F, and C may be found. 

Example. — Find the circle through the three points (o, o), (a, o), 

(0,6). 



We get by substitution, 
C =0 . . 


(1) 


a 2 -f- 2 Ga + C = . . 


(2) 


b 2 + 2 6F + C = . . 


(3) 



c = o 

F = - 



.-. the equation of the required circle is x 2 + y 2 = ax + by. Similarly 
the circle through any three given points may be found. 

57. Oblique axes. Equation of circle. — If in § 53 the 
axes are inclined at an angle <f>, the equation of the circle 
whose center is (h, k) and radius r, is, 

Yi 

P(x.y) 




Fig. 83. 

(x - hf + (1,- kf + 2 (x - h) (y - k) cos <f> = r 2 . (1) 

Expanding, the second-degree terms are^ 2 -f- y 2 -+- 2xy cos<£. 
. the most general form of the equation to a circle is 

A (x 2 + f + 2 xy cos $) -f 2 Gkc + 2 Fy + C = . (2) 

AYe may find its center thus : Divide through b}^ A, and 
compare coefficients with those of equation (1) above. 



THE CIRCLE 



89 



G F 

We get -£- = — h — k cos <f>, -r- = — k — h cos <f*, 

n 

— =h 2 + k 2 + 2 M cou £ - r 2 . 

Solving these three equations, we find h, k, and r. 

•Note. — The above equations are seldom, if ever, used In general, 
it is easier to study curves from their equations referred to rectangulrr 
axes. Oblique axes give rise to very complicated formula? which ari 
useless for practical purposes. These formulas are derived in text-books 
for the sake of completeness. 

58. Definitions of tangent, normal, subtangent, sub- 
normal. — If a secant P Q of any curve revolve about one of 
the points of intersection P 
until the point Q moves up to 
coincidence with P, the limit- 
ing position PT of the secant 
PQ is called a tangent to the 
curve at P. The line PS _L to 
the tangent PT at P is called 
the normal to the curve at P. 
The distance MN is called the 
subtangent. and MR the sub- 
normal. 




Fig. 84. 



Note 
teacher. 



— The idea of tangency should be clearly explained by the 



59. Equation of the tangent to the circle x* 



r 



?~ at 



the point (x v y x ). — Let P be (x l ,y 1 ), and Q an adjacent 
point on the circle (x 2 , y 2 ). 

y - Vx = Vi - y 2 



Equation of chord PQ is 



a) 



If Q approaches P, then ultimately y x 



- & = 0, x x - x z = 0, 



90 ANALYTICAL GEOMETRY 

and the dexter of (1) assumes the indeterminate form - • The 

difficulty is removed by the consideration that P and Q 
always are on the circle. 

... a . i « + y i . = t J (2 ) 

^2 2 + 2/2 2 = ^ (3) 

or, (x* - xf)+(y* - yf) =* (4) 

whence (x t - x 2 ) (x t + x 2 ) = - (i/ 1 - y 2 ) (i/ 1 + y 2 ). 

. Vx - y-2 = x, + x 2 
x x - x 2 y t + y 2 

(1) becomes .-. l^Zll + .?L+3 = 0. ...... (6) 

x — x l y 1 + y 2 

'which is the equation of secant PQ. 

This will become a tangent at P if x 1 = x 2 , y 1 = y 2 . 
Making this substitution, and clearing of fractions, we get 

Wi - Vx + xx x - x x = °> 

or, xx x + yy t = x* + y* == t* 

.'. the required equation of the tangent at (x 1 ,y 1 )is 

xx x + Wx = **■ 

EXERCISES. 

Find the equations of the following circles : 

1. Center (4, — 5), radius 3. 

2. Center (1, 2), radius 7. 

3. Center (— 5, — 3), radius 5. 

4. Center (1, 0), radius 2. 

5. Center (0, 0), radius 4. 

6. Center (5, — 1), radius 1. 

7. Center (0, 2), radius f. 

8. Center (3, - 3), radius 6. 



THE CIRCLE 91 



9. Center (h, k), radius Vh 2 + k 2 . Ans. x 2 + y 2 = 2 hx + 2ky. 

10. Distance between (1, — 3) and (3, 5) as diameter. 

Ans. x 2 + y 2 — 4 x — 2 y — 12 = 0. 

11. In the first quadrant, touching both axes, and radius = r. 

Ans. x 2 + y 2 ' — 2 rx — 2 ry + r 2 = 0. 
Find the centers and radii of the following circles : 

12. x 1 + y 2 + 12 x - 6 y = 0. 

13. x 2 + y 2 - 3 x - 9 = 0. 



\2a 2 a / 2 a 



+ C. 



15. 


x 2 + ?/ 2 — 4x + 4?/ — 1 = 0. 


16. 


4 (x 1 + y 2 ) + 12 ax -6 ay - d' 


17. 


(x + y) 2 + (x-y) 2 -8a 2 = 0. 


18. 


sec (x 2 + y 2 ) — 2 ax — 2 ay U 



I 3a 3a\ la. 
-4ns. (0, 0) ; 2 a. 



,4ns. (a cos 0, a sin 0); a. 

19. x 2 + y 2 + 2 x + 2 y + 1 = 0. 

20. 3 (x 2 + y 2 ) - 8 x - 6 y + 4 = 0. 

21. Show that the circle x 2 + y 2 + 2x + 2y + 1 = touches both 
axes, and find the points of contact. Ans. (— 1, 0), (0, — 1). 

22. Find the equation of the circle through the points (2, 3), (1, — 4), 
(2, - 5). 

23. Through (1, 2), (4, - 5), (- 1, -1). 

24. Through (0, 0), (a, 6), and (6, a). 

Ans. (a + 6) (x 2 + y 2 ) - (a 2 + b 2 ) (x + y) =0. 

25. Through (1, 2) and (4, 5) and having a radius 8. 

26. Find the equation of the circle whose center is on the line 

Sx + 4y — 7 = 0, 
and which touches the lines 

x + y - 3 = 0, 
x- y - 3 = 0. 

^4ns. 9x 2 + 9y 2 - 42x + 47 = 0. 

27. Find the circle which touches both axes and passes through the 
point (2, 6). 

28. Show that the equation 9x 2 + 9y 2 - 42 x 4- 38 y - 59 = rep- 
resents a circle. 



92 ANALYTICAL GEOMETRY 

Find the points of intersection of the following pairs of curves : 

29. x 2 + y 2 = 9, x 4- y - 1 = 0. 

30. x 2 + y 2 = 6 x, x = y 

31. x 2 + y 2 - 2x - Sy + 3 = • 0, x - y 4- 1 = 0. Ans. (1, 2), .(£, §). 

32. x 2 + V s - 2x - 4y - 3 = 0, x + y - 7 = 0. Ans. (3, 4) (3, 4). 
Hence the line touches the © at that point, 

33. Find the circle which touches both axes at a distance of 5 from 
the origin. Ans. x- 4- y 2 — 10 x - 10 y + 25 = 0. 

34. Show that the circles, x 2 4- y 2 — 6 x — 6 y -f 10 = 0, and x 2 +■ ?/ 2 = 2, 
touch each other, aud find the point of contact. Ans (I, 1). 

35. A line of fixed length always has its extremities in the axes. 
Show that its mid-point describes a circle 

36. Find the equations of the tangents to the circle x 2 + y 2 = 10 at the 
points whose abscissa is 1. Ans. x ^ 3 y = 10. 

37. Two lines are drawn from (a, o) and ( =- <v o) to make an angle $ 
with each other. Show that the locus of its vertex is 

x 2 -\- y 2 — a 2 = ^-2 ay ctn <p. 

38. Given base = 2 m, and sum of squares of sides = 2 S 2 [in a A], .find 
locus of vertex. Ans. x 2 -f- y 2 + m 2 — S 2 = 0. 

39. Given [in a A] base = 2m, vertical angle = 0, find the locus -of 
the intersection of the altitudes on the two other sides. 

Ans. x 2 + y 2 + 2 my ctn <p - m 2 = 0. 

40. Find the equation of the tangent from the origin to the circle 

x 2 + y 2 — 3x + 4?/ = 0. Ans. 4y-3x = 0. 

41. - Find the equatiomof -the- tangent to the circle x 2 4- y 2 — 25 at the 
point (— 3, 4). 

42. Find the radius of the circle x 2 + y 2 — 6x — 8y4-25 = 

Ans. 0. 

43. For what value of k will the line y = x + k touch the circle 
x 2 + y 2 = 4 ? 

44. Find a tangent to the circle x 2 4- y 2 = 8x and parallel to the line 
2x4-32/4-4 = 0. 

45. Also a tangent to the circle x 2 4- y 2 = 9 and _L to the line 
x-y-1 = 0. 

46. A tangent to the circle x 2 4- y 2 - 10 x — 8y + 2 = and parallel to 
the linex — 2y = 0. 



THE CIRCLE 93 

47. Show that the line y = x + a\/2 always touches the circle 
x? + y 2 = a 2 . 

48. For what value of k will the line 2x — 3y + k =0 touch the circle 
x 2 + y 2 - 8x-8y + 12 = 0? 

49. Find the circle through (2, 3), (4, 5), and (6, 1). 

Ans. 3 (x 2 + y 2 ) — 26 x - 16 y + 61 = 0. 

50. Find the circle on the distance between (x v ?/,), (x 2 , y 2 ) as a 
diameter. Ans. (x — x x ) (x — x. 2 ) + (y — y x ) (y — y 2 ) = 0. 

51. Show that the line 4x + 3 y — 35 = touches the circle 

x 2 + 2/ ? - 2 x - 4 y - 20 = 0. 

52. Find the length of the chord which the circle x 2 + y 2 — 5x — 6 y + 
6 = intercepts on the x-axis. .<4ns. 1. 

53. Find the tangent to the circle x 2 + y 2 — k 2 which makes a A of 
area k 2 with the axes. Ans. x + y = k V2. 

54. Find the chord made by the line y — x — 3 = on the circle 
x 2 + y 2 - 2x - 2 y - 23 = 0. Ans. \/82. 

55. Show that the line & (y — 6) — ax = touches the circle x (x — a) 
+ y (y — b) =0, and find point of contact. Ans. (o, 6). 

56. Find the chord of the circle x 2 + y 2 — 2x — 3^ + 3 = made by 
the line x — y+ 1 = 0. 1 

V2 

57. A (a, o), B (— a, o), are two points. Find the locus of P such 

(n 2 + 1 \ 
ln^l a '°j - "n 2 -l 

58. A point P moves so that the sum of the squares of its distances 
from the sides of a square (side = a) is constant = 2 k 2 . Find its locus. 
Take two adjacent sides of square as axes. 

Ans. x 2 + y 2 — ax — ay +a 2 — k 2 = 0. 

59. Find the circle through (2, 3), (3, 4), (0, 0). 

Ans. x 2 + y 2 - 23 x + 11 y = 0. 

60. Find the locus of a point which moves so that the sum of the 
squares of its distances from the sides of an equilateral A = k 2 . 

Let the vertices be (— a, o), (a, o), (o, a \/3). 

Ans. 6 (x 2 + y 2 ) + 6 a 2 - 4 (k 2 + ay \/3) = 0. 

61. Find a tangent to the circle x 2 + y 2 — 4x — Gy + 12 = and 
parallel to 'Sy — 4x + 2=0. 



that PB = n • PA. . . . . (n 2 + 1 \ .. 2 na 
Ans. A circle, center —z a, o) , radius 



94 ANALYTICAL GEOMETRY 

Suggestion. — S.y — 4 x + k = is the tangent required where k is to 
be found from the condition that the distance from center of to the 
line = radius. 

62. Find tangent from origin to (-) x 2 + y 2 — Sx — 8 ?/ + 5 = 0. 

Ans. y = kx where k is determined as above. 

Note. — The tangent in Ex. 61 may also be found by eliminating y and 
finding k, so that the points of intersection are coincident. 

63. Three tangents to the circle x 2 + y 1 = 25 form an equilateral A, 
and one of them is parallel to the x-axis. Find their equations. 

fy+5 = 0. 
Ans. \y-x / 5-x-l0 = 0. 
[y + Vb-x-10=fO. 

60. Equation of tangent at (x v y x ) to the circle x 2 + if 

+ 2 Gx + 2 Fy + C = 0. — The reasoning here is similar to 
that of the preceding article. 
The equation of line PQ is 

y - Vx _ Vx - y* m 

Also, ^ + ^ + 2G^ + 2F yi + C = . . . (2) 
* 2 2 + yi + 2 Gx 2 + 2 ¥y 2 + C = . . . (3) 
By subtraction, 
<X - x 2 ) (x t ■fx 2 + 2G) + ( 2 / ! - y 2 ) (y t + y, + 2 F) = 0, 

Vx — y* _ ^1 + ^2 4- 2 Gr 



or 



»! - «a i/i + 2/ 2 + ^ F 



/-•si V — Vi x. + x„ ■+■ 2 G- 
.♦. (1) becomes ? — * = 1 n ^ .... (4) 

W x -x t y 1 + y 2 + 2F w 

Now put # 2 = x v y 2 = y v and the equation to the tangent 
becomes 

y-Vx = _ g i + G 

a; - a 1 y t + F 

Clear of fractions, (x — x x ) (x 1 -f- G) + (y — y x ) (y 4 + F) = 0, 



THE CIRCLE 95 

■*■ xxt. + yy x + Gx + Fy = £ t 2 + ^ + Gs x + Fy r 
Add Gx x -f- Fy t + C to both sides, giving, 

**i + Wi + G{x + x ± ) + F (y + yj + C 

= *x 2 + ^i 2 + 2 G^ + 2 Fy t + C 
= [since (a^, y x ) is on O]. 

,\ the equation of the tangent at (x v y t ) is 

a», + yft -f G (a3 + ajj + F (y + y t ) + C = 0. 

61. The Normals. — In § 59 the equation of the normal 
[a _L to the tangent at (x v yj\ is 

y - lh = x - x i 
Ux x i 
or, x k y — y x x = (1) 

which shows that the normal passes through the origin. 

Otherwise, its equation is y — y t = — (x — x t ). 

x i 

[x 1/1 

Slope of tangent is , hence slope of normal is ^ • 

This equation also reduces to x t y — y t x = 0. 

V -h F 

In § 60 the normal is y — y t = l (x — x x ). 

X^ -j- Lt 

[x +G ?/ +F1 

Slope of tansrent is — =, , hence slope of normal is — — — \. 
y,+^ F ^-hGj 

The equation reduces to 

* (Pi + F ) - V ( x i + G) + (h/ t - Yx, = 0, 

which shows that the normal passes through the center. 

These normals may also be found by assuming the pre- 
viously known fact that they pass through the center. 

Thus, for O x 2 *f if = r 2 , the normal is the line joining the 
points (0, 0) and (x v y t ), i.e., 



96 



ANALYTICAL GEOMETRY 







x _ y :1 

x x y x 1 
1 




= 0. 


For the circle of §60, the line joining 
- G, - F), viz., 

x y 1 

*i Vi 1 
-G -E 1 


the points (x 1? yj and 
= 0. 



From these, the equations of the tangents may be found 
(lines _L to these at x v y x ). 

62. Tangent to the circle (x - hf + (y — kf = r 2 at point 

(x v y L ). — If the circle be given in this form, the method of 
procedure is the same as in the two other cases. 



kf = i* 

kf = r 2 



(1) 

(2) 
(3) 



Equation of PQ is 

Then also (x 1 - hf + (y t 
(x 2 - hf + (y 2 
By subtraction, 

(x 2 - x ± ) (x 2 -f- x t - 2h) + (y 2 - y t ) (y 2 + y t -2k) = 0. 

Hence the secant becomes, from (1) above, 

y — y y _ _ Xo -\-x l — 2 h 
x — x t ~ y 2 -\- y 1 — 2k' 
Now put x 2 = x v y 2 = y v and reduce and simplify, getting, 
as the required equation of the tangent, 

(x, - h) (x - h) + (y t - k) (y-k) = r> . . (4) 
The normal is found to be 

(x - x t ) {y t - k) -(y- y x ) (x t - h) = . . (5) 

63. Equation of circle through three given points. 
Determinant form. — Let the points be (x v y t ), (x 2 , y 2 ), and 

\ x 3> yy* 



THE CIRCLE 



97 



Let the required circle be 

x 2 + y 2 + 2Gx + 2¥y + C = 0. 
Then x 2 + y 2 + 2Gx t + 2¥y t + C = 



X Z + y? + 



(1) 

(2) 
3 2 + y? + ♦ . ■ . . . etc. = ... (3) 

Eliminating G,F,and C, we obtain the required equation, viz., 
x 2 + y 2 & y 1 

1 



«a+ya, x 2 y<L 



+ V% 



Vz 1 



. =0 

etc. = 



= 



64. Condition for four concyclic points. — If the four 
points (x v y t ), (x 2 , y 2 ), . . . etc., lie on a circle, its equation 
is that found for any three of them ; viz., 



x 2 -f- y 2 x 
x a ~r y$ X 8 

x ? + y? a* 



y 
y* 
y% 
y* 



= o. 



Now, if the fourth point (x v y t ) lie on this circle, we have 



«i* + V? 



x 2 + y? 

r» 2 _|_ n , 2 



'2 
3+2/3 



iC 2 

x% 



y% 

y± 



= 0, 



x? + y? 

(is the condition that the four points be concyclic. 

Note. — This result may also be obtained thus : By elementary 
geometry, the sum of the products of the opposite sides of a cyclic quad- 
rilateral is equal to the product of the diagonals. Take for vertices 

(*V Hi) («2 2/2) » etC ' 

65. Equation of tangent in terms of slope. — Let the 

line y = mx + b intersect the circle x 2 + y 2 = y 2 . Then, the 



98 ANALYTICAL GEOMETRY 

condition that it may touch the circle is the condition that it 
intersect it in two coincident points. 

Eliminate y, x 2 + (mx + &) 2 = r 2 , 

or, (1 + m 2 ) x 2 + 2 bmx + b 2 — r 2 = 0. 

This quadratic, which determines the abscissae of the 
points of intersection, will be a perfect square, i. e., will have 
equal roots, if 

(1 + m 2 ) (b 2 - r 2 ) = m 2 b 2 
or, b 2 = r 2 (1 + ^ 2 )- 

.-. 6 = ± r Vl + m 2 . 
Hence the line y s= mx ± r Vl + m 2 

is a tangent to the circle for all values of m. 

Note. — Another method : If the line y = mx + b touches the circle, 
its distance from the center (0, 0) is equal to the radius. 

b 



±VT + 



. •. b = -[- r vl + m 2 , as before. 

66. If the equation to the circle be given in the form 

x 2 + y 2 + 2 Gx -f 2 ~Fy + C = 0, we find, in the same way, 
the equation of the tangent in terms of its slope ; viz., 

y + F = m (x + G) ± VG 2 + F 2 - C • VI + w 2 . 

67. Length of the tangent from (x v y t ) to the circle 

(x - hf + (y - kf = r 2 . 

_ : 1 .^ Hence, the square 

of the length of the 

Fig. 85. 



PT 2 


= PC 2 


— r 2 , 


PC 2 


= (*1~ 

+0a- 


- A) 2 
-A-) 2 , 


PT 2 


= (*i- 


- 7?,) 2 



THE CIRCLE 99 

tangent is the result of substituting the co-ordinates of the 
given point in the sinister of the circle when the dexter is 
zero, i. e., when the terms are transposed to one side. 

If the equation be given in the form 

x 2 + if 4 2 Gx 4- 2 Ty + C = 0, 
we obtain PC 2 = fa + G) 2 + fa + F) 2 , 

CT 2 = G 2 4 F 2 - C. 

.-. PT 2 = fa + G) 2 4 fa 4 F) 2 - (G 2 + F 2 - C), 
or, PT 2 = x? + y* + 2 Gx, + 2 Fy, + C, 

which agrees with the rule given above. 

68. Equation of tangent to the circle x 2 + if = ?* 2 from a 
given external point fa , y t ). — (Let x 2 , y 2 ) be any other point 
on either tangent (since two may be drawn). Then the 
equation of the line joining the two points is, 

x fa - 2/ 2 ) - V ( x i - x i) + x xV<t - 2/1*2 = 0. 
The condition that this line should touch the circle is that 
its distance from the center be equal to the radius. 

. X \Vl ~ V\ X 2 _ 

' ± ^fa ~ I/2)' 2 + ( x i ~ X -J 2 
Now, writing x, y, instead of x 2 , y 2 , we get the required 
equation, viz., 

i*\(x - xj- 4- (y - Vl f\ = \xy x - x,y\ 2 * 

Another method : In practice, it will be found more expe- 
dient to work thus : 

Let xx 2 4 yy, = r 2 be the tangent at the point of contact 
fa, ?/ 2 ) whose co-ordinates are undetermined. Then, since it 
passes through (x v y t ), we have, 



* For another method see the appendix. 



100 ANALYTICAL GEOMETRY 

^2 + 2/i2/ 2 = ?' 2 (1) 

and xi+y} = r> (2) 

since (x 2 , y 2 ) is on O. 

From (1) and (2) we get values for x 2 , y 2 , and substitute in 
the first equation taken. 



EXERCISES. 

1. If x 2 + xy + y 2 + 2x + 3 y = represents a circle, show that the 
axes make an angle = 60°. Also center is (— i, — |), and radius = 

2. Find equations of tangent and normal to x- + y 2 = 25 at the point 
(3,-4). ^ins. 3*-4y = 25, 4x + 3?/=0. 

3. Find the circle whose center is (1, — 2), radius 5, the axes mak- 
ing angle <f> = 120°. Ans. x 2 — xy + y 2 — 4 x + 5 y — 18 = 0. 

4. Find the tangent at the origin to the circle x 2 + y 2 + hx + ky = 

Ans. hx + ky = 0. 

5. Find the tangent at (4, 5) to the circle x 2 + y 2 — 6x + Sy — 
12 = 0. 

6. Find the tangent at the origin to the circle x-+ y 2 + 2x + Sy = 

Ans. 2x + Sy = 0. 

7. Find the tangents to z 2 + y 2 = 4 which are inclined to the cc-axi& 
at (1) 45°, (2) 30°, (3) 60°, (4) 120°, (5) 150°. 

8. Find the tangents to x 2 + y 2 = 36 which are inclined to the x- 
axis at 60°. Ans. y = x V3 + 12, y = x Vs - 12. 
Also, those at 45°. Ans. x — y-\-V2 = 0. 

9. The axes making an angle 0, find the equation of the circle 
through the origin making intercepts a, 6, on the axes. 

Ans. x 2 + 2 xy cos 4> + y 2 = ax + fo/. 

10. Find the point of contact of the line x + y = 2 + V2 which 
touches the circle x 2 + y 2 — 2x — 2?/+l = 0. 

J.ns. (1 + f V% 1 + ^-\/2). 

11. Find the length of the tangent from (3, 4) to x 2 + y 2 — 22 x + 16 y 
+ 4 = 0. 

12. Find the length of the tangent from (1, — 2) to 3 x 2 + 3 ?/ 2 — x — 
4 = 0. Ans. \A£. 



THE CIRCLE 101 

13. Find the locus of a point such that if tangents be drawn from it 
to two concentric circles, their lengths may have the ratio k : 1 ; also m : n. 

14. Find the length of tangent from the origin to x 2 + y 2 — 4 x — y 

+ 2 = 0. 

15. Find the condition that - -j- j- =1 may touch the circle x 2 + y 2 =r 2 . 

Ans. a 2 b 2 = r 2 (a* + 6 2 ). 

16. Find the equation of the tangent from (6, 8) to the circle x 2 +y 2 =25. 
Take y = mx-^-r Vl + m 2 . 

Then x, y, r, are known. Solve for m, etc. 
Find the equation of the tangent to the circle. 

17. x 2 + y 2 = 25, whose slope is 2. 

18. x 2 + y 2 = 9, whose slope is |. 

19. x 2 + y 2 = 4, which is H to y — 2 x — 1 = 0. 

20. x 2 + ?/ 2 = 1, which is ± to 2x — y + 2 = 0, also which makes 
60° with x-axis. 

21. x 2 + y 2 - 2 x - 2 ?/ + 1 = and ± to 2 x + 3 y + 4 = 0. 

22. x 2 + y 2 - 8 x - 6 y + 12 = and || to 2 x - 5 ?y - 9 = 0. 

23. Find the equation of the tangent to x 2 + y 2 = 2 rx at (x n y { ). 

24. Find the length of the tangent from (8, 10) to the circle x 2 + y 2 
-3x = 0. Ans. 2v 7 35. 

25. Find the equations of the tangents to the circle x 2 + if — 4x — 
2 y + 1 = which are || to the x-axis. Also those drawn to it from 
the origin. Ans. y = — 1, y = 3, x = 0, 3x + 4y=0. 

69. Equation of the chord of contact for the point (x t , yj* 

— Let (x 2 , ?/ 2 ), (x s , y 3 ), be the points of contact of the tangents. 
Then these tangents are 

. xx 2 + yy<t = r 2 (l) 

xxt + m=-**. ( 2 ) 

But (x l} y^) lies on each tangent ; hence, 

*x x * + y^y* = i* ( 3 ) 

^1^3+2/l2/3 = ^ (4) 

* Definition. — The chord of contact of any point with respect to 
a given circle is the line joining the points of contact of the tangents 
drawn to the circle from that point. 



102 ANALYTICAL GEOMETRY 

From equations (4) and (5) it is seen that the co-ordinates 
of both the points (x 2 , y 2 ), (x s , ?/ 8 ), satisfy the equation 

a»i + Wi = ^ ( 5 ) 

Hence this is the required equation, that of the chord of 
contact. 

If the circle is 

* 2 + f + 2 Gx + 2 Fy + C = 0, 

by similar reasoning its chord of contact is found to be 

™>i + VVx + G (« + *x) + F (y + //O + C = 0. 

Special Case. — If (x v y } ) is on the circle, equation (5) represents a 
tangent at that point. Hence the chord of contact of a point on the 
circle coincides with the tangent at that point. 

70. Poles and Polars. — Def. If a chord of a circle turns 
about a fixed point in its plane, the locus of the point of inter- 
section of the two tangents drawn at the extremities of the 
chord is called the polar of the given point. 

The fixed point is called the pole of that locus (with re- 
spect to the circle). These definitions for pole and polar are 
the same for the other curves which we shall study. * 

71. To find the polar of the point (x v y x ) with respect to 
the circle x° + if = r 2 . — Let (as 2 ,y 2 ) be the point of inter- 
section of the two tangents. We must now find the locus of 
the point (x 2 , y 2 ). 

The chord of contact of this point is 

xx 2 + yy 2 = r 2 (1 ) 

But, by definition, it passes through the fixed point (x lf y t ), 
•'•x 1 x 2 + y 1 y 2 = r & (2) 

* The teacher should explain these conceptions, using illustrations. 



THE CIRCLE 103 

Equation (2) shows that the co-ordinates of (x 2 , y 2 ) always 
satisfy the equation 

xx t + yy l = r2 ( 3 ) 

which must therefore be the locus of (x 2 ,y 2 ), i.e., the polar of 
the point (x ti y x ). 

The polar is evidently a straight line. 

Special Cases. (1) If (x p y y ) is on the circle, equation (3) is also a 
tangent. Hence the polar and tangent of a given point on the circle are 
identical. 

(2) If (x,2/,) is an external point, (3) represents its chord of contact ; 
hence the polar and chord of contact of an external point are identical. 

72. Theorem. — If the polar of P t passes through P 2 , then 
the polar of P 2 passes through P r 
Let P 1 be (x v y t ), P 2 (x 2 , ;/ 2 ). 
Then the polars of these points are 

* x x + Wi = r * (!) 

**2 + Wfa = ^ ( 2 ) 

Now, if (1) passes through P 2 , we have 

* 2 *i + VtVi = 7 ' 2 (3) 

and if (2) passes through P p we have 

*i* 2 + ViVz = >' 2 (4) 

But (3) and (4) are identical ; hence, etc. 




^tei.Vi) 




^x(xi.yi) 



r A{£C2,y 2 ) 



Fig. 87. 



104 ANALYTICAL GEOMETRY 

73. Theorem. If the polar s of P t and P 2 intersect in P, 
then P is the pole of the line P x P 2 . %t 

Proof : The polar of P t passes through P, 

.'. the polar of P passes through P r [§72.] 

Similarly, the polar of P passes through P 2 . 

.-. P X P 2 is the polar of P, or P is the pole of PjP 2 . Q.E.D. 

74. To find the pole of the line Ax + By + C = with 
respect to the circle x 2 + if — r 2 . 

Let (x v y x ) be the pole required. 

Then its polar is xx 1 + yy 1 = ?- 2 . 

ISTow if this equation and the given equation represent the 
same line, they must differ only by a constant factor, i.e., the 
coefficients in both are proportional. 

• ?! _ Hi = _ r 2 

' * A B C* 

Ar 2 Br 2 

These values are real if A, B, C. are real. Hence every 
real line has a pole. 

Note. — By reasoning, similar to that employed in preceding articles, 
the chord of contact of the circle x 2 + y 2 -f- 2 Gx + 2 Fy + C = is found 
to be, xx, + yy x + G (x + x x ) + F (y + y x ) + C = 0. 

The polar of (x v y,) is the same equation, and the previous 
remarks hold true here. 

75. Geometrical construction of the polar and pole. — Let 

P be the given point whose polar is required. Take center 
as origin, and OP as cc-axis. . Then the equation of the circle 
is x 2 + y 1 — r 2 , and P is the point (x x , 6). 



THE CIRCLE 



105 



^2 

The polar ,of P is xx x = r 2 , .'. x = — , which is a line par- 

x i 

allel to the y-axis and at a distance — from it. Hence, to 
a x. 




Fig. 88. 

construct it, find B so that OP X OB = OA 2 , i.e., a third 
proportional to OP and OA (r). Then MN _L to OP at B is 
the polar required. 

To find the pole of a given line MN, draw OB ± to it, and 
determine P so that OB x OP = OA 2 = r 2 . 

EXERCISES. 

1. Find the polars of (1, 4), (- 2, - 5), (3, - 1), (0, 1), (- 1, - 2), 
(— 4, 1), with respect to the circle x 2 + y 2 =? 49. 

2. Find the poles of x - y + 1 = 0, Sx + 4 y - 6 = 0,. 

? + F«l y- 6 x +12 = 0, 22/-4x-5 = 0, 
h k 
with respect to the circle x 2 + V 2 — ^5- 

3. Find the pole of x cos a + ?y sin a = p with respect to the circle 

x 2 +y 2 = 20. 

20 cos a 20 sin a 



Ans. 



P 



106 



ANALYTICAL GEOMETRY 



4. Show that the nine-points' circle of the A whose vertices are 
(2 a, - J , (2 6, - J , (2 c, - J , passes through the origin. 

5. Prove that the distances of two points from the center of a circle 
are proportional to the distances of each from the polar of the other. 

6. Find the condition that the polar of (h, k) with respect to the circle 
x 2 + y 2 = 9 may touch the circle x 2 + y 2 = 6 y. Ans. h 2 + 6 k = 9. 

7. If the polars of the vertices of a A ABC form a A A'B'C, prove 
AA', BB', CC, are concurrent. 

Note. — The polars of the vertices of A'B'C are the sides of ABC. 

8. If a circle is inscribed in a A, show that 
the lines drawn from the vertices to the oppo- 
site points of contact are concurrent. 

Suggestion. — A ABC and A'B'C bear 
the same relation to each other as those in 
Ex. 7. 

9. Eind the condition that the polar of 
(h, k) with respect to x 2 + y 2 = a 2 may touch 

(x - a) 2 + (y - p) 2 = r 2 . 




Fig. 89. 



Ans. (ah + pk- a 2 ) 2 = r 2 Qi 2 4- A: 2 ). 



10. If a circle is inscribed in a quadrilateral, 
show that the diagonals and the lines joining 
the opposite points of contact are all concur- 
rent. 

11. Eind the polar of (— 9, — 7) with respect 
to x 2 + y 2 = 14. 

12. Eind the polar of (— 1, — 2) with respect 
to x 2 + y 1 = 3. 

13. Eind the pole of 2»+6?/+14 = with 
respect to the circle x 2 + y = 5. 

14. Pole ofx — y — 2 = with respect to x 2 + y 2 = 8. 

15. Pole of ax + by — 1 = with respect to x 2 + V 2 = 49. 

Ans. (49 a, 49b). 

16. Eind the points of contact of the tangents from (4, 7) to the circle 

x 2 + y 2 = 64. 




Fig. 90. 



THE CIRCLE 107 

17. Show analytically that the angle inscribed in a semicircle is a 
right angle. 

18. Under what conditions will the circle Ax 2 + Ay 2 + Dx + E?/ + 
F = touch the axes ? Arts. D 2 = 4 AF, E 2 = 4 AF. 

19. Find the condition that the circles 

x 2 + y 1 + 2 G x x + 2 F t y = 0, ) may touch each other 
x 2 + y 1 + 2 G 2 x + 2 F 2 y = 0, j at the origin. 

. G. G 2 

Ans f;=f 2 - 

76. The radical axis of two circles.* — Let the two circles 

hp 

x 2 + f + 2 Gx + 2 % + C = . . . (1) 

and x 2 + if + 2 G^ + 2 F l2 / + C x = . . (2) 

Then the equation 
x 2 + f + 2Gx + 2Fy + C = x 2 + f+2G x x + 2F 1 ?/+C 1 (3) 

passes through the points of intersection, real or imaginary, 
of (1) and (2) [why ?], and represents a straight line, since its 
equation reduces to 

2(G-G 1 )« + 2(P~r i )jf + C-C 1 = 0. . (4) 



* Note 1. — One of the circles may be a "point-circle." Thus the 
radical axis of 

(x - h) 2 + (y - k) 2 - r 2 = (1) 

and (x - h,) 2 + (y - &J 2 =0 (2) 

is the line 

2 (/^ - h) x + 2 (A;, - k) y + h 2 + k 2 - h 2 - k 2 - r 2 = . (3) 

It is the. locus of a point which moves so that its distance from a given 
point (h v k x ) is always equal to the tangent drawn from it (the moving 
point) to the circle (1). 

Note 2. — The equation Sj — S 2 =0 [ § 76 ] shows that the square of 
the length of the tangent from (x, y) to Sj = is equal to the square of 
the tangent from (x, y) to S 2 = 0. 

Hence, the radical axis of two circles is the locus of a point from which 
tangents drawn to the two circles are equal. 



108 



ANALYTICAL GEOMKTRY 



It is called the radical axis of the two circles. 
For brevity, if the © be S x = 0, S 2 = 0, the Sj - S 2 = 
represents their radical axis. 

77. The line of centers of two circles is perpendicular to 
their radical axis. Its equation may be written. 



x y 

- G - F 



{} i ~ F i 1 



0, 



or x (Fj - F) - y (Qt t - G) + GF, - YG l = 0. 

This is evidently I to the radical axis, equation (4), §70. 

78. Common tangent of two circles. — If two circles are 
tangent to each other, their radical axis is their common 
tangent at the point of contact. 

Example. Determine A so that the circles 

x i + 7 f _ 6 x - 4 y + A. = . . . . (1) 
a j* + 3/ 3 + 4a-3y + A = .... (2) 

may touch each Other. 

Suggestion. Find radical axis by subtraction, then determine X so 
that the radical axis may intersect either in two coincident points. 

79. Radical center of three circles. — Let S = 0, Sj = 0, 
S a = 0, be three circles. Then the radical axes of these 
circles taken in pairs are 

S _ s x = 0, S - 8 a = 0, S, - S a = 0. 

Now, any values of X and // which satisfy any two of these 
equations, evidently satisfy the third. Hence the radical 

axes of three circles taken in pairs are concurrent in a point, 
called the radical center of the three circles. 

Note. Tangents drawn from the radical center to the three circles 
arc all equal. 



THE CIRCLE 109 

80. Coaxial Circles. — The equation S — AS X = 0, repre- 
sents a circle passing through, the points of intersection of the 
© S = and S x = 0. It may represent any one of a system 
of circles according as we assign different values to the vari- 
able parameter A. Such a system passing through two fixed 
points (which may be real or imaginary) is called a coaxial 
system. The line joining these two points is the common 
radical axis of any two circles of the system. 

Note 1 . — S — XS 1 = shows that the tangents to the two circles 
S = 0, S t = from any point, are in a constant ratio X : 1. Hence, if a 
point moves so that the tangents drawn from it to two given circles are 
in a constant ratio, its locus is a coaxial circle. 

Note 2. — If S = be one circle of a coaxial system whose radical 
axis is Ax + By + C = 0, then any other circle of the system may be rep- 
resented by the equation S + k (Ax + By + C) =0 where k is some 
absolute constant. For the radical axis of the two circles is 
S - I S + k ( Ax + By + C) } = 0, or Ax + By + C = 0, the given radi- 
cal axis. Hence . . . . , etc. 

81. To represent a coaxial system, having given two 
fixed points A and B and the distance 2 c between them. — 
Take AB as y-axis, and O, its mid-point, as origin. 

Then the cc-axis, or the line 
x = 0, will be the common • ^a a (o.c) 



radical axis of the system. // 

Also, x 2 + y 2 — c 2 = repre- / // 

sents one of the circles, — the / / ' 

-+-H- 



\ 



circle whose center is the 

origin. \w 

\ V 



.'.by note 2, § 80, the equa- 
tion ^1b( ,-c) 
x 2 + if-c 2 -\-2kx = Q . (1) fig- 9i. 

represents any circle of the system where c 2 is constant for all 
the circles while k is a number which varies for each circle. 



110 



ANALYTICAL GEOMETRY 



Note. — It is evident that the centers of the system will lie on the axis 
of x. The system may .-.be represented also by the equation (x — X) 2 
+ y 2 = c 2 where X varies with each circle and is the distance of its center 
from the origin. 

The equation (1) above may be written in the form 

(x + k) 2 + y 2 - k 2 + c 2 . 

Now, if k 2 -f- c 2 = 0, this becomes a " point-circle." 

.-. k = ± c V— 1. 

.*. the circle becomes one of the points (± c V— 1, 0). 
These points are called the limiting points of the system. If 
the © intersect in real points, c is real, and the limiting points 
are therefore imaginary ; but if the circles intersect in 
imaginary points ± a V— 1, then there are two real limiting 
points (± a, o)* 

82. The angle of intersection between two circles. — 

This is the angle between the tangents, one to each circle, 
at their point of intersection. 

Tangents at P, the point 







■^. 


of intersection of the two 




ScV 




circles, are _L to the radii 




fp\U 




\ AP and BP. Hence the 


1 '/ [ 

1 4_ 1 


1 \\\ 

L 4 N * 




\ angle between the tangents 


V ^ "' 


1 "B 




1 is equal to angle APB. 
/ Let AB = 8, 
AP = r 1 , 




Fig. 92. 




BP = r 2 . 


Then 


S 2 = r* 


1 + r} - 


- 2r 1 r 2 cos <j>. 




.'. COS 


r' 

rk 1 


2 + r 2 - 8 2 




9 — 


2r x r 2 



* See note 7, appendix. 



THE CIRCLE 



111 



If the (D be given, 

x 2 + ^ + 2G 1 x + 2F 1 y + C 1 = 
^ 2 + / + 2 G. 2 x + 2F 2 y+C 2 = 



and 
Then 



(1) 

(2) 



' = G x 2 + F x 2 - C t , r 2 2 = G 2 2 + *7 - C 2 , 

.-.^(^-G^+^-F,) 2 . 

.-. 2 r t r 2 cos <£ = 2 G t G 2 + 2 F^ - C t - C 2 . 

Note. — The condition that the two circles cut orthogonally is 
2 G^a + 2 FjF 2 - C, - C 2 = 0. 

83. To find the circle which cuts orthogonally the three 
given circles. 

z 2 + 2/ 2 +2G 1 * + 2F 1 ?/ + C 1 = 

tf -f tf + 2 G 2 a + 2 F 2 y + C 2 = 

^ + ^ + 2G,a+ 2.F 8 y + C, = 

Let the required circle be 

z 2 + y* + 2 G* + 2Fy + C = 0. . . 

Then 2 GG X + 2 FF 1 - C - C x = . . 

2 GG 2 + 2 FF 2 - C - C 2 = . . 

2 GG 3 + 2 FF 3 - C - C 3 = . . 



(1) 
(2) 
(3) 
(4) 



Eliminating G, F, and C from (1), (2), and (3), and (4) 
we get, 

x 2 + f x y 1 



- c 2 
-a 



G 2 



F x -1 



0, 



which is the required equation of the circle cutting the three 
given circles at right angles.* 



* See notes 5 and 6, appendix. 



112 ANALYTICAL GEOMETRY 



DIAMETERS. 



84. Definition. — The diameter of a curve is the locus of 
the mid-points of a system of parallel chords. 

85. To find the equation of a diameter of the circle 

x 2 + if = r 2 . — Let any one of a system of parallel chords be 
y = mx -f b. 

Also let it meet the circle in the points (x t , y{), (x 2 , y 2 ). 

Then m = - x * + * 2 . [ See § 19, (1). 1 

2/i + y 2 L J 

Also, if (x, y) are the co-ordinates of its middle point, we 
have, 

2 y = Vx+y<L- 

X 

.'. m = — -y 

y 
1 

or V = x . 

J m 

is the required equation of the diameter. 

Discussion. — This equation shows that the diameter passes through 
the center of the circle and is perpendicular to its chords. 



EXERCISES. 

1. Find the equation of the circle, center on ?/-axis, and passing 
through (0, 0) and (x v y y ). Ans. y x (x 2 + y 2 ) = y {x 2 + y 2 ). 

2. Find equation of the circle which passes through (a, b) and touches 
the line - + y = 1 at the point (3 a, - 2 b). 

3. A secant to the circle x 2 -f y 1 = r 2 is drawn through the fixed point 
(x v y^. Find the locus of the mid-point of the chord. 

Ans. x 2 + y 2 — x x x — y x y = 0. 



THE CIRCLE 113 

4. Find the length of the common chord of the ©s. 

(x - b) 2 + (y - a) 2 = r 2 . ) v 

Find the condition that these circles touch. 
Suggestion. — Put the result = 0. 

5. For what point on the circle x 2 + y 2 = 4 is the subtangent equal to 
the subnormal ? Also for what point has the tangent a slope = 1 ? 

6. Through a fixed point (x v y v ) on the circle x 2 + if = r 2 a chord is 
drawn. Find locus of its mid-point. 

Arts, (x - | x,) 2 + (y - |y x ) 2 = (J r) 2 . 

7. x 2 4- ^/ 2 = r, 2 , x 2 -f ?/ 2 = r 2 2 , are two concentric circles. Find the 
locus of a point from which tangents to the. circles are inversely propor- 
tional to the radii. Ans. x 2 4- if — r 2 + r, 2 . 

8. A chord of the circle x 2 4- y 2 = 49 is parallel to the line x — y = 0, 
and is 6 units long. Find its equation. 

Suggestion. — Take x — y + fc = 0. Find points of intersection 
with circle. Then express that chord = 6. Find k. 

9. Find radical center of circles 

a 2 + 2/ 2 + 2x 4- 2 ?y - 1 = 0, I ulns. (3, 2). 

z 2 + y 2 + 3x = 0.J 

10. Find radical axis of the circles 

* + , -8,-6,+ 12 = 0,) ^ 8»-M«--7. 

x 2 4-y 2 4-4x-9?/4- o = 0. ) 

11. Radical axis of 

z 2 + y 2 - 2z + Sy = 1, 
x 2 + y 2 4- 4 x — 5 y = 5. 

12. Radical center of 

z 2 + y 2 - 6x4- 2 = 0, 

x 2 4- y 2 4- 4 y 4- 1 = 0, 

SB 2 + y 2 - hx -> 9y 4- 15 = 0. 

13. Show that the circles 

x 2 4 y 2 -I- 14x4- r oy - 7 =0,1 

x 2 4- y 2 4- 4x+ 2/ — 3 = 0, [ are coaxial. 

z 2 + y 2 ~ x - i/ - 1 = 0, J 



114 



ANALYTICAL GEOMETRY 



14. Find the length of the common chord of the circles 

x 2 + y 1 - 5 x - 12 y + 3 = 0, 

x 2 + y 2 + 4x - 9y + 6 = 0. 

15. Find the circle through (5, — 9) and the points of intersection of 
the circles 

x 2 + y 2 - 2 x - 2 y + 1 = 0, 
x ? + y 2 + 8x - 6y + 4 = 0. 

16. Find the circle through the origin and the points of intersection of 

x 2 + y 2 - 8 x - 0, 
x 2 + y 2 + 4y — x — 2 = 0. 

17. Find locus of a point such that tangents from it to the circles 

x 2 + y 2 + 4y= 0, 
x 2 + y 2 = 9, 

18. Find the circle through the origin and the points of intersection of 



are in ratio 4:7. 



x 2 + y 1 + 4z 



= 0, 



3x + 4?/ + 2 = 0. 
19. Find the circle whose diameter is the common chord of the circles 
x 2 + y 2 - 23 x + 11 y = 0, 
x 2 + y 2 - 12 x + 11 =0. 

Ans. x 2 + if — 5 x — 7 y + 18 = 0. 

86. Centers of Similitude. — The line of centers O'O of two 
circles is divided harmonically * at A and B in the ratio of 




Fig. 93. 

their radii; the points A and B are called the internal and ex- 
ternal centers of similitude, respectively. Several properties 



Harmonically means internally and externally in the same ratio. 



THE CIRCLE 115 

of these points are proven by elementary geometry, which the 
student can readily show analytically ; viz., 

(a) The external common tangents of the circles pass 
through B ; the internal tangents through A. 

(/?) Any line passing through either center of similitude is 
cut similarly by the two circles, thus : 

BP : BP' :: BQ : BQ':: r : /. 

(y) The six centers of similitude of three circles, taken ii, 
pairs, lie on four straight lines called the axes of similitude. 

87. Common Tangents to Two Circles. — To find the tan- 
gent to two circles, we must ascertain, 

(1) The centers* of the circles. 

(2) A center of similitude (x v ?/ t ). 

(3) For what value of m, y—y 1 = m(x — x^) will touch 
either one of the circles. 

Example 1. Given the circles, 

x 2 + if- - 10 x + 12?/ + 25 = (1) 

and x 1 + y 1 + 8 x = (2) 

The centers are (5, — 6), (— 4, 0), radii are 6 and 4 ; their ratio is §. 
Hence, the internal center of similitude is, 

3(- 4) + 2(5) 3 • + 2(- 6) 

Xl = 3~T2 ' V * = ST2~ ~ ' 

and the external center is, 

_ 3 (-4) -2 (5) _ 3 • -2(-6) 

X ' 2 ~ 3-2 ' %2 ~ 3^2 

.-. the external center is (— 22, 12). 

. \ y — 12 = m (x + 22) is the equation of any line through that center. 

This will touch the first circle if its distance from the center (5,-6) is 

equal to its radius 6. 

-18 -27 m . , . , , 

6, whence in may be found. 



=kvT 



+ m< 



116 ANALYTICAL GEOMETRY 

Ex. 2. Find the common external tangent to the circles 

X 2 -u y 2 = 16 (1) 

x 2 + y 2 + 6x-Sy = (2) 

The line y = mx -J- 4 \/ 1 + ?n 2 is a tangent to (1) for all values of m. 
Now treat this equation and (2) as simultaneous, and determine m so 
that they may intersect in two coincident points. 



EXERCISES. 

1. Find the external common tangents to the circles 

x 2 + y 2 -4x + 6y-9 = 0. 
x 2 + y-Q X + 10 ?/ + 4 = 0. 

2. Find the internal common tangents to the circles 

x 2 + y 2 - 9 y - 15 = 0, 
X 2 + y i + 6 x - 3 y = 0. 

3. Find the external and internal common tangents to the circles 

x 2 + y 2 - 8 x + x 6 y - 14 = 0, 
x 2 + 2/ 2 -2z-2?/+ 1 = 0. 

4. Show that the tangents from the origin to the circle 

(x - a) 2 + (y - b) 2 = r 2 
Also touch the circle 

( X _ fca) 2 + (y - kb) 2 = (kr) 2 . 

5. Find the circle through the origin and the points of intersection of 
the circles 

x 2 + y 2 + 2G 1 x + 2¥ 1 y+ C, =*0, 
x 2 + y 2 + 2 G 2 x + 2 F 2 y + C 2 - 0. 
^4ns. (C,-C,) (x 2 + 2/ 2 )+2(G 1 C 2 -G 2 C 1 )x + 2(F 1 C 2 -F 2 C 1 )?/=0. 

6. A point moves so that the square of its distance from the base of 
an isosceles A = product of its distances from sides. Find its locus. 

Let vertices be (— a, o), (a, o), (o, b). 

Ans. b (x 2 + y") + 2 a 2 y — ba 2 = 

88. Polar Equations of the Circle. — Let be the pole, 
ON the polar axis. 




117 



O N 

Fig. 94. 

P is any point on the circle, and C is the center. 
Now, in A OPC, 

CP 2 = OP 2 + OC 2 - 2 OP • OC cos POC. 
... p * - 2 PPi cos (0-0J+ Pl * - >*=0 . . . (1) 

This is the mcst general equation of the O in polar co- 
ordinates. 

Special Cases. (1) If ON coincide with OC, then e l = 0. 

.-. p 2 - 2 pp, cos 6 + /V - r 2 = (2) 

(2) If the pole is on the circle, p 1 » r. 

.-. p - 2rcos {6 - 6 X ) = (3) 

(3) If pole is on the circle, and polar axis passes through the center, 

p x =r, 1= O. 
.-. p - 2r cos0 = (4) 

(4) If center is at pole, p, = 0. 
.-.By equation (1) above, we get, 

P = r (5) 

r 
I /O I I / j I 

Case I Case II Case III Case IV 

Fig. 95. Fig. 96. Fig. 97. Fig. 98 

The Special Cases. 




118 ANALYTICAL GEOMETRY 

EXERCISES. 

1. The diameter of the through (o, o), (p 1? 0J, (p 2 , 2 ), 

is Vft 2 + p 2 2 - 2 /^/o, cos (d l - 2 ) + sin (d 1 - 8 2 ). 

2. The condition that the line 

1 
P 



a cos + b sin 
may touch the circle p = 2 a cos 
is a 2 5 2 + 2 a& - 1 - 0. 

3. The tangents from the origin to the circle (x — a) 2 + (y — 6) 2 = r 2 
are (6x — ay) 2 = r 2 (x 2 + ?/ 2 ). 

4. If AB is the diameter of a circle, the polar of A with respect to 
any circle which cuts the first orthogonally, passes through B. 

5. One vertex of a rectangle is fixed (h, k), two others move on the 
circle x 2 + y 2 = r 2 . Show that the fourth moves on the circle 

x 2 + y 2 + h 2 + k 2 - 2 r 2 - 0. 

6. The locus of the center of a circle which cuts two fixed circles at 
right angles is their radical axis. 

7. The centers of three ©s are A, B, C, and radii R v R 2 , R 3 . Show 
that the circles are coaxial if 

BC • R x 2 + CA • R 2 2 + AB ■ R 3 2 + BC • AB • CA = 0. 

8. The common tangents to the circles 

x 2 + y 2 — 2 ax = 0, 
x 2 + y 2 - 2 by = 0, 

are represented by the equation 

2 ab (x 2 + y 2 - 2 ax) = \ab - ax + by] 2 . 

NoTE . __ T he equation p 2 - 2 pp x cos (0 - X ) + p 2 - r 2 = is a 
quadratic in p. Thus, for each value of 0, there are two values of p ; 
viz., OP and OP r 

The product of the roots is p 2 - r 2 , 
i.e., OP • OP, = OC 2 - CP 2 

= (OC - CP) (OC + CP), 

Which agrees with a well-known theorem in elementary geometry. 



THE CIRCLE 



119 



89. Polar Equation of the Circle x 2 + y 2 + 2 Gx + 2 F+y 

= 0. 

Take the origin for the pole. 

.-. x = p cos 0, y — p sin 0. 

These give p 2 + 2 P (G cos + F sin 0) + C = 
for the required equation. 

If the pole is on the circle, i.e., the circle passes through 
the origin, then C = 0, and the equation becomes 

p -f 2 (G cos 6 + F sin (9) = 0. 



EXERCISES ON CHAPTER V. 

1. Given the base whose extremities are (x v y t ), (x 2 , y 2 ), and the 
vertical angle \f/ of a A, find the locus of the vertex. 

A ns. (x - x x ) (x - x 2 ) + (y - yj (?/ - ?/,) = -J- rtn \p x y 1 

*, 2/, 1 
z 2 ?./ 2 1 

2. The centers of three circles are (a,, 6,), (a,, 6 2 ), (a 3 , 6 3 ), and their 
radii are r, , r 2 , r 3 . If these circles are taken in pairs, show that the 
three external centers of similitude lie on the line whose equation is 

r, r, r, 
6 I 6 2 b 3 
a x a 2 a., 

3. A, B, C, etc., are given points. Find locus of P such that 



r, 


r 2 


»*3 




6, 


& 2 


K 


X — 


1 


1 


1 





r. 


r 2 


r. 




«i 


a 2 


a 3 


2/ = 


1 


1 


1 





PA + PB' + PC +,etc, 
Also, ra, PA 2 + m., PB ? + m 3 PC 2 +, etc., 



a constant. 



a constant. 

Ans. Circles. 

4. (a, o), (— a, o), are the ends of base of a A, ^ the vertical angle, 
dhow that the locus of the orthocenter is 

x 2 + y 2 + 2 ay ctn \f/ — a 2 == 0. 

5. Two tangents from (ft, k) and their chord of contact to the circle 

\ a l (fti _L ^2 _ a 2W § 

x 2 + y 2 = a 2 form a A. Show that its area is i il_. 

ft 2 + V 



120 ANALYTICAL GEOMETRY 

6. Circles are described on the three diagonals of a complete quadri- 
lateral as diameters. Show that they have, two by two, the same radical 
axis. 

7. Given four points A,B,C,D and four circles are described through 
the points taken in groups of three, viz., AJtiC, ABD, ACD, BCD, 
Show that any two of them intersect at the same angle as the remaining 
two. 

8. Given four straight lines, which intersect, and taken, three at a 
time, form four triangles. Show that their orthocenters are collinear. 

9. The circle whose diameter is the distance between (p t , X ) and 

(Pa , 2 ) is 

p 2 = p x p cos (0 - X ) + p,P cos (0 - 2 ) - p,/o 2 cos (0 l - 2 ). 

Note. — Let A and B be the given points, P (p, 0) any point on the 
circle. . •. PA 2 + PB 2 = AB 2 , etc. 

10. The locus of a point whose polar with respect to x 2 + y 2 = a 2 
touches (x — h) 2 + (y — k) 2 = r 2 , is 

(hx + ky - a 2 ) 2 = r 2 (x 2 + y 2 ). 

11. Two points, A and B, are on rectangular axes, origin O, so that 
OA = OB = a. Find the locus of a point P so that /_ OP A = /_ OPB. 

Ans. (x — y) (x 2 + y 2 — ax — ay) = 0. 
What does this last equation represent ? 

12. A,B,C, are three points on a circle, O any other point on it. 
From O ±s are dropped on the sides of A ABC. Show that their feet 
are collinear. 

Note. — Use polar co-ordinates, O as pole, and equation of © 
p = 2 a cos 0. 

13. If the circles (y — b) 2 + (x — a x ) (x — cl, ) = 0, 

0-B) 2 + (3J.-AJ (x-A 2 ) =0, 
touch each other, prove either 

(B - b) 2 + (A, - a,) (A 2 - a 2 ) = 0, 
or (B - b) 2 -f (Aj - a ) (A 2 - a,) = 0. 

14. S, = and S 2 = 0, are two circles, radii r x and r 2 . The circle 
whose diameter is the distance between their centers of similitude is 

5i — ^i = o 

r 2 r 2 
It is called the circle of similitude of the two ©s. 



THE CIRCLE . 121 

15. Tangents to these ©s from any point on their © of similitude are 
in the constant ratio of their radii. 

Note. — This follows from Ex. 14. 

16. The locus of a point P at which the ©s subtend equal angles is 
the © of similitude. 

17. Show that the ©s 

S S 

— - zL — = cut orthogonally. 

t 1 r 2 

18. Show that the ©s of Ex. 17 bisect the augles between the circles 
s, = 0, S 2 = 0. 

19. S, = 0, S 2 = 0, are two circles ; r, , r 2 , their radii. 

Show that their external and internal centers of similitude are re- 
spectively the centers of the ©s. 



s, 




s. 


= o, 


r, 




r 2 


S, 




S 2 




«■ 


+ 




= o, 



'1 '2 

and .*. the six centers of similitude of three ©s lie, three by three, on 
four straight lines. 

20. Find the locus of the centers of circles, which, viewed from two 
fixed points, subtend constant angles. 

21. Also, the locus of centers of circles which cut each of two given 
[fixed] circles in the extremities of diameters. 

22. Find the locus of a point such that the chords of contact of the 
tangents drawn from it to three given circles may be concurrent. 

23. Find the locus of a point such that the sum of its distances from 
two or more given straight lines is constant. 

24. From a fixed point P tangents are drawn to two ©s which pass 
through two fixed points. Find the locus of the intersection of the chord 
of contact with the diameter through P [in each ©]. 

25. A, B, C, D, are four concyclic points, O any other point. 
Prove : 

OA 2 • area BCD + OC 2 • area ABD - OB 2 • area ACD - OD 2 • area 
ABC = 0. 

26. Show that the radical axis of a circle and a point bisects the dis- 
tance between the point and its polar with respect to the circle. 



122 . ANALYTICAL GEOMETRY 

27. T x , T 2 , T 3 , are the lengths of the tangents from any point to 
three coaxial circles whose centers are A, B, C. 

Prove : BC • T, 2 + CA • T 2 2 + AB • T 3 2 = 0. 

28. A circle can be inscribed in the quadrilateral formed by the axes 

and the lines 

x cos a + y sin a = p x , \ 



x cos p + y sin p = p 2 
p x [sin p + cos p + 1] = p., [sin a + cos a + 1]. 

29. Two variable circles, which are tangent to each other, are also 
tangent to two given circles. Find the locus of the point of contact of 
the variable circles. 

30. A variable circle cuts two given circles at constant angles <p l , <f> 2 . 
Show that it cuts their radical axis at a constant angle \j/ determined by 
the equation d cos \p = r x cos t — r 2 cos <f> 2 . 

[fj , r 2 are the radii ; d, the distance between centers.] 

31. If R is the radius of the variable in the preceding exercise, and 
P is the ± from its center on the radical axis, prove : 

P = R cos yj/. 

32. A series of 0s are given, which, taken two by two, have the same 
radical axis. If a variable cuts two of these at constant angles, show 
that it will cut the remaining circles at constant angles. 

33. Prove that a circle which cuts two given circles at constant angles 
touches two fixed circles. 

34. On the axes a variable rectangle OABC is constructed with a 
given perimeter = 2 p. Prove that the ± from vertex C on diagonal AB 
always passes through a fixed point. 

35. Show that the center of the inscribed circle of the A whose 
vertices are (x x , y x ), (x 2 , y 2 ), (x 3 , y 3 ), 

a x, + bx 2 -f- ex., ay, + hy 2 + c y 3 J 
a + b + c a + b + c 

where a, b, c, are the lengths of sides of A. 

36. Two 0s touch each other internally, and a third touches both 
[one internally and one externally]. Show that the sum of the distances 
of the center of the third circle from the centers of the two given circles 
is constant. 



THE CIRCLE 123 

37. A variable circle cuts three given 0s at equal angles. Show that 
the locus of its center is a _L from the radical center on one of the axes 
of similitude. 

38. Find the circle which cuts the three circles 

x 2 + y 2 = a 2 , 1 * 
(x — b) 2 + y 2 =± a 2 , I orthogonally. 
x 2 + (y - c) 2 = a 2 , J 

Ans. x 2 + y 2 — bx — cy + a 2 = 0. 

39. A chord of a © moves parallel to itself, and lines are drawn 
through its extremities to meet at a given angle. Find the locus of their 
intersection. 

40. Find the system of 0s which cuts orthogonally each of the system 

x 2 + if + 2 \ x - a 2 = 
[where X is a variable parameter]. 

Ans. x 2 + y 2 + 2 X 2 y + a 2 = 0. 
[where X 2 is a variable parameter]. 

41. If the axes make an /_ = 60°, find the center and radius of the 
circle represented by the equation 

x 2 + xy + y 2 — 4 x — 5 y — 2 = 0. 
^4ns. (1, 2); 3. 

* The center of required is radical center ; radius = a tangent there- 
from to one of the 0s. 



CHAPTER VI 

TRANSFORMATION OF CO-ORDINATES 

90. The reference of a curve to a new set of axes 
sometimes simplifies its equation, and thereby facilitates 
the investigation of its properties. Of course, a curve remains 
unchanged by any transformation of co-ordinates, although its 
equation, whose nature depends in a measure on the relative 
position of the axes, may assume a slightly different form; 
and, as we shall see presently, the degree of an equation 
remains unchanged by any variation of axes. We shall dis- 
cuss only the most useful forms of transformation. 

91. To transfer the origin to the point (h, k) without 
changing the direction of the axes. 



r 




Y' 

P 


M 










(h,k) 


\n 

1 

1 T 


(0,0) 




" •**>• 

N L 




(P.o) 



Fig. 99. 



Fig. 100. 



OX, OY are the old axes ; O'X', O'Y' are the new axes. P 
is any point in the plane whose co-ordinates are (x, y) to the 
old axes (V, y') to the new axes. 

124 



TRANSFORMATION OF CO-ORDINATES 



125 



Now, x = OL = MO' + O'L' ' 

= A + a?', 
y = PL = NO'+PL' 
= * + y. 
Hence, to transform to a new origin (A, k) with new axes 
parallel to the old, we put x' + h for x, and y' + k for y; or, in 
the end, the accents may be dropped, giving 
x + h for x, 
y + k for ?/. 

Note. — To transfer back to the old axes, we put 
x — h for x, 
2/ — k for y. 

92. To transform from one set of rectangular axes to 
another, the origin remaining the same; i.e., to turn the 
axes through a given angle. 

r Y 




Fig. 101. 

P is any point (x, y) old axes, and (x', y') new axes. 
PM' is _L to OX' and M'M is _L to PL. 



Now 



Similarly, 



OL = OL' - LL' 

= OL' - MM' 

= OM' cos 6 - PM' sin 6. 
x' cos — y' sin 6. 
x' sin -f- y' cos 0. 



126 



ANALYTICAL GEOMETRY 



The accents may be dropped after transformation. 
The combined changes of §§91 and 92 are evidently ef- 
fected by the use of the formulae, 

x = h -f- x' cos 6 — y' sin 0, 
y = k + x' sin 6 + y' cos 0. 

93. To change from rectangular to oblique axes with 
the same origin. 

The positive directions of the new axes make the angles a 

and /3 respectively with the 
old a?-axis. 

These formulae are 
readily obtained : 

x — x' cos a + y' cos ft, 
y = x' sin a + y' sin (3. 

If wetake/?= 90° + a, 
these formulae reduce to 
those of § 92. 



7 






F'« 








\(3 


\cc 


p 

A 90-/3 

/Lx' 

' — (r1 l 
! ! 


—•^ 












X 



Fig. 102. 

The combined changes of §§91 and 93 are represented 
thus: 

Y 




Fig. 103. 

x = h + x' cos a + y' cos j3, 
y = Jc + x' sin a + ?/ sin /?. 



TRANSFORMATION OF CO-ORDINATES 



12; 



94. To transform from one set of oblique axes to 
another. 




Fig. 104. 

PM' is II to OY', PL and M'L' are II to OY, MM' is II to OX 
Now from A OI/M', 

LW s in ("), 



and 



L'M' = x' 



Sill co 
sin a 



Also, from A PMM', 



Sill u> 



MM'=/ sin ^-"> 



and 



MP = y 



sin (o 

, sin/? 
sin a 



XT AT AT' TV/TTV/T' , Sm(cO — a) , Sill (/? — to) 

Now a: = OL = OL — MM = x ^ — - y — 



sin <o 



sin a) 



-r -,» -r /Tit-/ ™^» , SHI CX , SUl /? 

y = LP = L'M' + MP = x' -. h y' 



sm a> sin to 



. sm (co — a) , sm (co — j3) 

,-. x — x — 7 + y — ^ — 9 

sin co sm co 

, sin a , sin B 

y = x' -. h y' ^-1- ■ 

sm co sm co 



128 ANALYTICAL GEOMETRY 

95. It has thus been seen that in the transformation 
of axes, the co-ordinates x and y are replaced by expres- 
sions of the fiirst degree: functions of a;', y', the new co- 
ordinates, and certain angles made by the axes with each 
other. Hence, the degree of an equation cannot be raised. 
Likewise it cannot be lowered, for in that case the degree of 
the new equation could be raised by reverting to the old axes. 
.-., etc. 

MISCELLANEOUS EXERCISES. 

1. M is the mid-point of side BC of A ABC. Prove analytically, 

AB 2 + AC 2 = 2 AM 2 + 2 BM ? . 

2. Show that the diagonals of the rectangle formed by the lines 

x = «i, ) y = &i, 

x = a 2 , ) y = & 2 , 

(6 t — b. 2 ) x — (a x — a 2 ) y + aft, — ab x = 0, 
(&! — b,) x + (a t — a 2 ) y — a,6, + ab 2 = 0, 

3. Menelaus' Theorem : If a straight line meets the sides of a A ABC 
in X, Y, Z, prove 

BX CY AZ 
XC ' YA ' ZB _ 
Suggestion. — Take A (x v y x ), B (x 2 , y 2 ), etc., and given line 
ax + by + c = 0, 
find values for the ratios. 

4. The equation 

2 Rxy +2Gx + 2Fy + c = 
represents two straight lines if 2 FG = CH. 

Show also that these lines form a ^7 with the axes. 

5. Show that the lines 

tan a = - ( bisect the angles between the lines 

x V represented by the equation 
and x cos a + y sin a = a ' 

(X - a cos a) 2 + (i/ — a sin a) 2 = fc ? jxcos a -f ysin a — aj 2 . 



TRANSFORMATION OF CO-ORDINATES 129 

6. Show that the diagonals of the ^7, Ex. 4, are 

F G U ' 

7. Determine k so that the equation 
k (x 2 + y 2 — a 2 ) + j x cos ( °^ — J + y sin ( — - — J — a cos f — - — j 

(*«-("-^) + »-»(^)-.«-f^)|-o 

may represent two straight lines. 

^ns. fc_ J8in(^)sin(^)j 



EXERCISES ON CHAPTER VI. 

Transform 

1. 4x — 5 y — 6 = to || axes through (2, — 5). 

2. x + 2 ?/ — 1 to || axes through (1, 4). 

3. x 2 + y 2 - .8 x + 6 y - 12 = to || axes through (— 1, - 3). 

4. 2 x 2 - 6 y 2 - 4 y + 5 x - 9 = to || axes through (- 3, 4). 

5. x - 2 y - 9 = to || axes through ( - 5, - 3). 
Turn the axes through 45° in the equations 

6. x 2 - y 2 = 4. Arts, xy + 2 = 0. 

7. 2y-3x+l = 0. 

8. 3x 2 - 4xy + 3?/ 2 + 24 = 0. 

9. x* + 2/ 4 + 6 xy -2 = 0. ^L?is. x 4 + ?/* - 1 = 0. 

11. Turn the axes through /_ in x 2 + y 2 = r 2 - 
"Why is there no change in the equation? 

12. Turn the axes through an angle = tan - 1 f in 2 x 2 — 5 xy + x - 4 = 0. 

13. Transform x 2 4- y 2 — 5 = to || axes through (1, 1). 

14. Transform x 2 + y 2 — 2 x = to |( axes through (1, 1). 

Ans. x 2 + y 2 + 2 y = 0. 



130 ANALYTICAL GEOMETRY 

x 2 y 2 

15. What does the equation - — ~ = 1 become, if the new axes 

make ^/s, tan - 1 -J- \ with the old axis, and origin is unchanged ? 

Ans. 4 xy = 5. 
Transform to polar co-ordinates, pole at origin, 

16. x 2 + y 2 = r 2 . 21. 2 x + 3 y - 1 = 0. 

17. x 2 + 2 ax - y - 12 = 0. 22. x - 2 y + 4 = 0. 

18. x 2 + ?/ 2 = 6 x. 23. x cos a + y sin a = p. 

19. ?/ = 3x. 24. - + ?= 1. 

a o 

20. x 2 - ?/ 2 = a 2 . 25. x - fey - a = 0. 

26. Transform x 2 + y 2 — 8x + 6 y — 14 = 0, pole at (1, £), polar axis 
|| to x-axis. 

27. Transform x 2 + y 2 + 4 x — 2 ?/ + 1 = 0, pole at ( — 2, — 5), polar 
axis at 30° to x-axis. 

28. Turn axes through 45° in 5x 2 + 6 xy + 5?/ 2 — 8 = [rectangular 
axes]. Ans. 4 x 2 -I- y 2 — 4 = 0. 

29. Also in x 2 — y 2 = 2. Ans. xy = — 1. 

30. In x 2 + 2 xy tan 20 — y 2 = 2 a ? , turn axes through ^/ 0. 

Ans. x 2 — y 2 = 2 a 2 cos 2 0. 

31. If, in turning a pair of rectangular axes through /_ 0, the equation 
becomes Ax 2 + 2 Bxy + By 2 = 

prove A'x' 2 + 2 K'x'y' + By 2 = 0, 

A' + B' = A + B, ) 
A'B' - H' 2 = AB - H 2 . J 
In the same transformation, prove 

x n + y n = x i + y \ 

Transform to rectangular co-ordinates, 

32. p 2 sin 29 = a 2 . 36. tan 2 d = ab. 
33.. p- cos 2 = 4. 37. sec = H. 

34. p = 5. 38. p = a sin cos (9. 

35. p = 4 sin 0. 39. p = - • 7 cos 2 0. 

a o 

Transform to polar co-ordinates, pole at origin, 

40. x 3 - 3 x 2 y - 3 xy 2 + if = 0. Ans. tan 3 = 1. 

41. mx 3 — 3 mxy 2 + y s — 3 x 2 y = 0- Ans. tan 3 = w. 



TRANSFORMATION OF CO-ORDINATES 131 

42. Transform to rectangular co-ordinates, p^ cos^ 6 = a*. 

43. Transform x 2 — 2xy + 2y 2 — 4x + 6?/=0to axes turned through 
tan —1 f . 

44. To what point must the origin be moved [axes parallel to old 
position] in order that the equation 

(a) 5 x 2 + 4 xy + 8 if - 18 x - 36 y + 9 = 
may have no terms of the first degree ? Ans. (1, 2). 

Suggestion. — Let (x', y') be the new origin. Transform, and equate 
coefficients of x and y to zero. Solve for x f , y\ etc. 

(/3) x?/ — ax — 6.y = 0. Ans. (6, a). 

45. Through what angle must the axes be turned in order to remove 
from the equation Ax + By + C = the x-term ? the ?/-term ? 

Ans. d = tan- 1 — -. 

d = tan- 1 — . 
A 

46. Through what angle should the axes be turned so that the xy — 
term in the equation 

4 x 2 — 24 xy + 11 y 2 + 20 = may disappear ? 

Ans. 6 = tan- 1 f. 

47. (x, y) are the co-ordinates of a point referred to rectangular axes. 
Find its co-ordinates referred to the lines 

x 2 y 2 . 

-j — j- 2 ±= 0, as axes. 

48. Determine so that the xy-term in 

x 2 + xy + y 2 — 3x — 3y = may vanish. 

^Ins. = 135°. 

49. Through what angle should the axes be turned so that x-axes may 
pass through (-4, 12) in 4x - Sy + 6 = ? 



CHAPTER VII 



THE PARABOLA 

96. Definition of Conic. — A conic section or a conic is the 

locus of a point which moves in a plane so that its distance 

from a fixed point called the 
focus bears a constant ratio 
to its distance from a fixed 
straight line called the direc- 
trix. This ratio is denoted 
by e or e : 1. Thus, if P is 
any point of the conic, P the 
focus, and DD' the directrix, 
we have 

Fig ' 105 - PP = e ■ PD always. 

The ratio e is called the eccentricity. 

When e = 1, the conic is called a Parabola. 

When e < 1, an Ellipse. 

When e > 1, an Hyperbola. 

Example 1. Find the equation of the parabola, focus at the point 
(2, 3), and directrix the line 3x + 4y — 2 = 0. 
Let (x, y) be any point of the required parabola. 
Its distance from the focus is V (x — 2) 2 -f (y — 3) 2 

3x -f 4y - 2 




and distance from the directrix is 



±5 



... (a; -2)2+ (y-SY 



/Sx + 4y - 2\ 2 . , 

f j— I is the required equation. 



Ex. 2. Find the equation of the ellipse focus (1, 1) and directrix 
2x — y — 1 = 0, eccentricity = |. 

132 



THE PARABOLA 133 

Here V (5 - If + (y - If = \ (^~ 1 ) 

is the required equation. 

Ex. 3. Find the equation of the hyperbola focus (2, 0), directrix 
x— 9=0, eccentricity §. 



V(x — 2) 2 + V 2 = | (x — 9) is the required equation. 

97. Equation of the parabola. Type-form. — Let F be the 

focus, DD' the directrix, and a _L to DD', drawn through 



P(x,y) 




Fig. 106. 

F, the aj-axis. Bisect EF in 0, which is called the vertex, and 
which by definition is a point on the curve. Let be the 
origin. 

Let OF == a [known distance]. 

Now PF = PM = EB, [by def.] 

PF 2 = PE 2 + FR 2 , or ER 2 = PR 2 + FR 2 . 
.-. (x + of = J/ 2 + (x — a) 2 , 

whence y 2 = 4 ax (1) 

is the required equation. 

Discussion. — Equation (1) shows 

(1) Curve is symmetrical with regard to cc-axis or its own axis. 

(2) Curve lies wholly on same side of y-axis (i.e., x always has the 
same sign as the constant a), since negative values of x give imaginary 
values for y. 



134 



ANALYTICAL GEOMETRY 



(3) When x = oo, y = oo, .-. curve is open, extending to the right 
without limit. 

Note 1. — Equation of directrix is x + a = 0. 
Note 2. — FP = EO + OR = x + a. 
Query. — What curves are these ? 



x 2 = 


4 ay, 


t = 


— 4 ax, 


x 2 = 


- 4 ay. 



98. Construction of parabola, focus and directrix given. — 

(1) By points. 

DI>' is directrix, F the focus. Draw EE _L to DD'. Bisect 
EF in 0, which is called the vertex, a point on the curve. At 



D 
C 






L 




/ / 
/ / 
/ i 
i i 
I / 
' 1 


E 





\ \ 
\ \ 
\ \ 
V \ 
X \ 
X \ 
X v \ 


K 


D 






•iS^ 



D 


A 


< 


^ r 






°( 


\ F 


-^■JL> 


Y 




°\ 


i' 






D 




\ 







Fig. 107. 



Fig. 108. 



any point K of EF (produced if necessary) erect the _L LL'. 
Now with F as center and EK as radius, cut this _L in Q and 
R. These are two points on the curve. For FQ = EK = CQ. 
In this manner any number of points may be found and joined 
by a continuous curve. 

(2) Mechanically. 

Place a ruler in coincidence with the directrix DD', and a 
right triangular ruler with one edge AC against the ruler. 
Take a string of length BC, fasten one end to B, and the other 



THE PARABOLA 135 

end to the focus F. Then slide A ABC along the edge of the 
ruler at directrix, keepiDg the string tightly pressed against 
the edge BC by a pencil-point P. This point will trace the 
required parabola. For PF = PC always. 

99. Internal and external points. — If C is any point 
(x, y) within a parabola [See Fig. 106, § 97] y 2 — 4 ax, we have 

y* _ 4 ax _ cE 2 - 4 a . OR [for the point C] 

= CR — PR , which is negative. 

Hence the y 2 — A, ax of an internal point is negative, and 
y 2 — A ax of an external point is positive, and equal to zero 
for a point on the curve. 

100. The latus rectum [the double ordinate through 
the focus] is a third proportional to any abscissa and its 
corresponding ordinate. 

In y 2 = 4 ax, put x = a, .-. 2 y = 4 a [= latus rectum]. 
Also, x : y :: y : 4 a. Hence, etc. 

101. The Squares of any two ordinates of a parabola 
are to each other as their corresponding abscissas. 

Let (x lt y x ) and (x 2 , y 2 ) be any two points on the curve. 
Then V\ — 4«^ 1? 

y 2 =Aax 2 . 
.'. Vx : y-i'- Xt : x 2 . .-., etc. 

EXERCISES. 

1. Is the point (3, — 2) outside or inside of y 2 = 6x ? 

2. The vertices of an equilateral A are on y 7 = 4 ax, one of them at 
the vertex. Show that its side has the length 8 a V3. 

3. At what point of \f- = 16 x is the ordinate twice the abscissa ? 

4. Find where the line ax + by = 1 cuts the parabola if = 4 ax. 

5. Find the equation of the line through the focus of y 1 = 4 ax and 
at 45° to x-axis. 



136 ANALYTICAL GEOMETRY 

6. A chord PP' passes through the fixed point K on the axis of a 
parabola. N and N' are the feet of ±s from P and P' on the axis. 
Prove ON ■ ON' = AK 2 [constant]. See Fig. 106, § 97. 

7. PFQ is a focal chord, PO meets the directrix in K. Prove QK is 
parallel to the axis of the parabola. 

8. Find the chord made by?/ — 2x — 1 = with y 2 = 12 x. 

9. Find co-ordinates of foci, equations of directrices, and lengths of 
latera-recta in the parabolas, 

(1) y 2 + 4 ax = 0. (3) (y - 4) 2 = 6 (x - 5). (5) y 2 = ax. 

(2) x 2 + 4 by = 0. (4) if + 6 x + 8 = 0. (6) x 2 = &y. 

10. Find the equation of the parabola through the origin, and (3, 4) 
its axis being the x-axis. 

Let y 2 = kx be equation required. 

Then 16 = &k, whence, k = -L 6 -, etc. 

11. The vertices of a A are three points on the parabola y 2 = 4 ax 
whose ordinates are y x , y 2 , y s . Prove, 

area of A = =- (y x - y 2 ) (y 2 - y 3 ) (y 3 - y,). 
o a 

12. Two chords through the vertex whose lengths are C, , C 2 , are ± to 
each other. Prove C,* Cf = 16 a 2 (C,* + Cf). 

13. Show that the line y = x + 2 touches y 2 = 8 x. Find the point 
of contact. 

102. Equation of tangent to the parabola at point 

(x v y t ). — We shall employ the secant method here as in the 
circle. It is applicable to any curve. 

Let (x 2 , y£) be an adjacent point. Then the line joining 
the two points is 

y - ?a _Vx-y, /-n 

*// i//-| 'J&~ 'J0c) 

Also, 





X 


— x x 




x, - 


x 2 


y? 


= 


4 ax A 


) 


y? = 


4 


y? 


— 


v? = 


4 


a (x 1 


— 




Vx 


- y< 




4& 





x 2 ). 

Vi + Vi' 



THE PARABOLA 137 



Hence (1) becomes, 

V - Vx 4 & 



x ~ x i Vi + 2/ 2 
or V Oi + y 2 )—Aax = y x 2 4 y x y 2 - 4 ax x 

= Vi ft [since y* = 4 ax x ] . . (2) 
Now put y*=yv x 2 = x v 

.-. 2 yy x — 4 ax = y x = 4 ax x . 

.-. jyi = 2 a (x+ ^) (3) 

is the required equation. 

Example. — Tangent at vertex (0, 0) is = 2 ax, or x = [i/-axis.] 

The normal is - — = — L , 

or 2 a (y - y x ) + y, (x - x x ) = (4) 

Note. — Equation (2) above, of the chord, is sometimes useful. 

103. Equation of parabola with origin (h, k). — If the 

vertex of the parabola [the origin] is removed to (h, k) with 

its axis parallel to the cc-axis, its equation is evidently 
(y - kf = 4 a (x - h). 
Different positions of the curves are indicated by these 

equations, 

(x-hy = ±±a(y-k) i 

(y - kf = ±±a(x- h), etc. 

which the student can easily interpret. 

104. Any equation of the form Ax 2 -f 2Gx + 2Yy 4- C = 0, 
or B?f 4- 2 Gx 4- 2 Fy 4- C = 0, represents a parabola. — We 
shall take a special example for illustration. 

Find focus, vertex, latus rectum, directrix, etc., of the 
curve 

7/4- 2y- 12x- 11 = 0. 

This may be written thus, 

y» + 2y + l = 12(3+l) > 

or, (y 4- 1) 2 = 4.3 (*+■!). 



138 ANALYTICAL GEOMETRY 

Hence, comparing with the type-form (y — k) 2 = ka(x — h), 
we see that the vertex is the point (— 1, — 1), the latus rec- 
tum is 4 a = 4.3 = 12. 

Since a = 3, and vertex is (— 1, — 1). .-. focus is (2, — 1) 
[add 3 to - 1]. 

The directrix is x = — 4, the axis of curve is y = — 1. 

Hence the curve has its axis parallel to the cc-axis and 
extends to the right. 

The two equations at the beginning of this section can, in a 
similar way, be shown to represent parabolas. 

105. Equation of tangent in terms of slope. — The 

abscissas of the points of intersection between the line 
y = mx + b and the parabola y 2 = 4 ax, are determined by the 

e( l uation {mx + bf = ±ax, 

or m 2 x 2 + (2bm-4:a)x-t-b 2 = . . . . (1) 

Its roots are equal, i.e., the line is a tangent if 

(bm -2af = b 2 m 2 . 

.'. 4 a 2 — 4 abm = 0, b = — 

m 

a 

222 

is a tangent to the parabola for all values of m. 

Again substituting this value of b in (1), it becomes 
a\ 2 „ a 



mx I = 0, or jc — 

mj m l 

[abscissa of point of contact]. 

Substituting in y 2 — 4 ax, 

2a 
we sret y = — . 

Hence the point of contact of the tangent is ( — — ] 
* 5 \m 2 > m) 



THE PARABOLA 139 

106. Normal in terms of slope. — Let ra x be the slope of 
the normal, then 

1 

TO, = • 

TO 

Again, the equation of the normal is 

(^-^)=-i( x -^) ^ 

Substituting m l for in this equation, we get 

y = mx — 2 am x — am^, 
which is the equation of the normal in terms of its own 
slope m v 

EXERCISES. 

1. Find a point on y 7 = 4 ax where the tangent is at 30° to x-axis. 

Ans. (| a, a\/3). 

2. The curves y 2 — ax, x 2 = by, meet at an angle = 

3 a$ b* 



tan— i 



2 (a* + 6 5 ) 



3. A tangent at P meets directrix in M. Prove PM subtends a right 
angle at the focus F. 

4. Find the tangents to y 2 = 4 ax, at 30° to x-axis. 

5. Find the equation of a parabola whose axis is parallel to the 
x-axis, and which is tangent to the line y — 2x-l = 0. 

6. A chord PQ of y 2 = 4 ax subtends a right angle at the vertex. 
Show that it passes through the fixed point (4 a, o). 

7. A O described on a focal chord as diameter is tangent to the 
directrix. 

8. The tangent at any point of a parabola meets the directrix and 
latus rectum produced at points equidistant from the focus. 

9. Find the focus and directrix of the parabola represented by the 
equation x 2 + 2 ax + 4 by + 4 be + a? = 0. 

Ans. (—a, — c — 6), y = b — c. 



140 ANALYTICAL GEOMETRY 

10. Find the equation of a tangent drawn from (9, — 14) to y 2 = 6 x. 

11. Find the equations of the tangent and normal to y 2 = 8 ax at 
(2, 4.) 

12. Show that the tangents at (x v y t ) and (x 2 , y 2 ) meet in the point 

13. The sum of the subtangent and subnormal for any point on y- = 
4 ax, equals | length of the focal chord parallel to the tangent at the 
given point. 

14. From any point on the latus rectum ±s are drawn to the tangents 
at its extremities. Show that the join of the feet of these ±s touches 
the parabola. 

15. N is foot of ± from focus on tangent at P. 
Prove FN 2 = FO ■ FP. 

16. The /_ between two tangents to a parabola = \ /. between the 
focal radii of the points of contact. 

17. A triangle is circumscribed about a parabola, i.e., is formed by 
three tangents. Prove, 

(1) The A formed by them is \ the area of the A formed by the 

points of contact. 

(2) Its orthocenter lies on the directrix of the parabola. 

(3) Its circumscribed circle passes through the focus. 

(4) The area of the A = — (p — p 2 ) (p 2 — p s ) (p 3 — p x ) where 

o a 

4a = latus rectum, and p v p 2 ,p 3 , are the ±s dropped from 
the points of contact on the axis. 

18. Find the locus of the center of an equilateral A formed by three 
tangents or three normals to a parabola. 

19. If ±s be dropped on a tangent to a parabola from two points on 
the axis equidistant from focus, the difference of their squares will be 
constant. If F lt P 2 , are the points, the product = 2 a • PiP 2 . 

20. The tangent at one extremity of a focal chord is parallel to the 
normal at the other extremity. 

Suggestion. — Let (x v ?/,), (.r 2 , y 2 ), be the extremities. 

Show the condition that the line joining them passes through the focus 
is y x y 2 + 4 a? = 0. 

Find tangents at the points, etc. 



THE PARABOLA 141 

107. Geometrical properties of the parabola. — 




Fig. 109. 

(1) The subtangent is bisected at the vertex. In 

VVx = 2 a ( x + x i)> 
put y = 0. 

.-. x = - x v i.e., ON = OT. 

(2) The subnormal is constant and equal to the semilatus 
rectum 2 a. In the equation of the normal, viz., 

V -¥!=-&(*-*& 

y = 0. 
x — x x — 2 a, 
NN'= 2 a. Hence, etc. 
= FT, for FP = CN = ON + a = OT + a = FT. 

.-. FP = FT. 
tangent bisects the angle FPM, 
= FT. .-. Z FPT = Z FTP = Z TPM. 

.-. Z FPM is bisected, etc. 
foot of al from the focus to a tangent lies in the 



put 

We get- 
or 

(3) FP 



-(4) The 
for FP 

(5) The 
i/-axis. 



142 ANALYTICAL GEOMETRY 

/ 

Let FM and PT meet in R. Then A PRF and PRM are 
equal. (Why ?) 

.-. ZPK,F = ZPKM. 

.-. R is the foot of the _L from focus to the tangent. 

Also, since FR = MR, and FO = OC, 

.-. OR is parallel to CM. X, etc. 

(6) If the tangents at Q and Q' meet in T, then a parallel 
to the axis through T bisects QQ'. (The student should draw 
the figure.) 

Let Q and Q' be (x v ?/ t ) and (x 2 , y 2 ) respectively. 

Then the equations of QT and Q'T are 

yy 1 =2a(x + x 1 ) (1) 

yy 2 = 2 a (x + x 2 ) (2) 

Solve for the co-ordinates of T. 

By subtraction, 

VQ/i -1/2) = 2a(x 1 -x 2 ) 

••• y = i (yi + y a ), 

which is the equation of the parallel to the ;r-axis through T, 
It evidently passes through the mid-point of QQ'. .-., etc. 

Note. — The co-ordinates of the point of contact of the tangent [§105] 
may be found thus : 

The tangent yy l = 2 a (x + x,) may be written 







2o 

y = — 


[ 2 ax, 

-x + 1 • 

2/i 


Comparing 


this with 


y = 


a 

mx -\ 5 

m 


we have 




2a 

m = 

V\ 


2a 


and 




a 2 ax, 
m~ y x ' 


2 ax, 
= — — l - = mx x . 
2a l 

m 






.'.X, 


a 



THE PARABOLA 143 

. \ point of contact of the tangent y = mx -\ is ( — ^ , — ) • 

m \m 2 m ) 

108. Tangents to a parabola from a given point. — 

y = mx -f- — is the tangent for all values of m. If it pass 
through any point (x v y x ), we have 





Vi 


V\ 


± 


a 

»! + -' 

m 






m 


Vtf- 


4 


ax x 






2*. 







These values are (1) real and unequal, (2) equal, or (3) im- 
aginary, according as yf — Aax l >, =, or < 0. .*. two dis- 
tinct, two coincident, or no tangents can be drawn from a 
given point to a parabola according as that point is outside, 
on, or within the parabola. 

109. Normals from a given point. — The equation to the 
normal in terms of its own slope m x is, 

y = m x x — 2 am x — am x z . 

« 

There are, in general, three roots of this cubic in m„ some 
of which may be imaginary. However, at least one normal 
can be drawn to a parabola from a given point. 

110. To find the locus of the foot of a perpendicular 
from the focus to a tangent. — Any tangent is 

a ... 

?/ = mx -\ (1) 

m K J 



y=-~(x-a) (2) 



A ± to it from focus (a, o) is 

1 
V " m 
Eliminate m from (1) and (2) by subtraction, 

.-. [m -\ — )x = 0, 
\ W 

.-. x = [the tangent at vertex]. .-., etc. 



144 ANALYTICAL GEOMETRY 

111. Locus of intersection of two tangents _L to each 
other. 

Let the tangents be 

y = mx + - (1) 

a 
y = m x x H (2) 

.-. mm x = — 1 (3) 

[Eliminate m from (1), (2), (3).] 

Subtract (2) from (1) and divide by (m — m^. 

r\ a 

.-. (J = x = x -f- a. 

.: x 4- a = [the directrix] is the required locus, etc. 

112. Angle between two tangents from (x v y t ). 
The tangent y = mx -\ 

passes through (x v y x ) if y x = mx x -\ , 

or m 2 x l — my t + a = (1) 

This equation shows that generally two tangents can be 

drawn from a given point. [See § 108.] 

Let m and m i be the roots. Then the tangents are 

a a 

y = mx -\ , y = m.x -\ • 

m m l 

If <f> is the angle between them, 

tan <f> = 3- • 

1 -f- «i% 

_ , . . ?/, a- 

From (1), m + m x — — , mw t = — 

.*. (m — w x ) 2 = (bi + ^i) 2 — 4 mm t = — — - • 

Vyi 2 — 4 o t 

tan d> = — ; . 

a + x x 



THE PARABOLA 145 

EXERCISES. 

1. Find the locus of a point from which the sum of the squares of 
the normals drawn to a parabola = k 2 [a constant] . 

2. The locus of the foot of a X from focus on a normal is the par- 
abola y 2 = a (x — a). 

3. The tangent at the extremity of the latus rectum is twice as far 
from focus as from vertex. 

4. A parabola moves parallel to itself so that the vertex traces the 
parabola in its original position. Tangents are drawn from vertex of 
fixed parabola to the movable parabola. Find locus of the points of 
contact. 

5. The circle on any focal radius as diameter touches the tangent 
[to the parabola] at the vertex. 

6. Tangents are drawn to y 2 = 4 ax at points whose abscissae are in 
the ratio r : 1. Show that the locus of their intersection is the parabola 
yi = (r* + r - *) 2 ax. 

7. Find the equations of the parabola referred to (1) the axis 
and directrix, (2) axis and latus rectum. 

8. Tangents y = m 1 x-\ 5 y = m 2 x -\ i meet in the point 

( — + 



; m l m 2 \ m 1 

9. The JL from the intersection of 

a 

y = m, x -f — 
v m 1 

y = m 2 x H 

m 2 . 

a 

on y = m,x -\ 

* m 3 

meets the directrix in the point 

\—a,a (—+ — + — + ) • 

( \m 1 7ft, m 3 m 1 ■ m 2 • mj) 

10. Tangents to a parabola at M and M' meet in P. 

VM 2 PUT'* 

Prove, tf!l. = t^L [where F is focus]. 

MF M'F L J 



* The symmetry of this result shows that the orthocenter of a A cir- 
cumscribed to the parabola lies on the directrix. 



146 ANALYTICAL GEOMETRY 

113. Diameters. — Let any chord of a parallel system in a 
parabola be y = mx + b, and let (x v y^) and (x 2 , y 2 ) be its ex- 
tremities. 

Then, if (x, y) is its mid-point, we have 

2x = x 1 + x z , 2y = y x + y 2 . 

. 4 a 4 a 2 a 
Again, m == - = 7r - = 

2a 
m 

is the equation of the diameter bisecting the chords || to 
y = mx -f- &. 

Discussion. (1) Every diameter is a straight line parallel to axis of 
the parabola. 

(2) Any line parallel to the axis is a diameter, for m and 2 a may have 
any value. 

(3) The tangent drawn through the end of a diameter is parallel to 
the chords of that diameter. 

EXERCISES. 

1. The ± from the focus to the chord y = mx + b meets the diameter 
of this chord in the directrix. 

2. The ± from the focus to a tangent and the diameter through the 
point of contact meet in the directrix. 

3. The tangents through the extremities of a focal chord intersect on 
the directrix and at right angles. 

4. The axis of the parabola is the only diameter which is perpendic- 
ular to the tangent at its extremity. 

5. A diameter lies above or below the axis according as its chords 
make an acute or obtuse angle (positive) with the axis. 

114. The chord of contact of the point (x x , y{). — Let 
( x 2f 2/2)? (^3* Vs)) b e the points of contact of the tangents drawn 
to the parabola from (x lf y x ). 



THE PARABOLA 147 

Then the tangents are, 

yy* = 2 a (x + x 2 ) (1) 

Wz = 2 a ( x + x s) (2) 

But both these pass through (x lt y x ) 

.*. ya/ 2 = 2 a Oi + * 2 ) (3) 

and 2/^3 = 2 a (^ + « 8 ) (4) 

From (3) and (4) it is evident that both the points (x 2 , y 2 ), 
( x 3> 2/3)5 li e on the line whose equation is 

yy 1 = 2a(x + x 1 ) (5) 

which is the required equation of the chord of contact. 

If (x v y t ) is on the curve, (5) is also a tangent at that point. 
.•. the chord of contact of a point on the parabola is coinci- 
dent with the tangent at that point. 

115. Polar of (x v y t ) with respect to the parabola y 2 = 

4 ax. — Let the tangents at the ends of the chord revolving 
about the fixed point (x u y^ intersect in (x 2 , y 2 ). 

To find the locus of point (x 2 , y 2 ). 

The chord of contact of this point is 

yy 2 = 2 a (x + x 2 ) (1) 

and since it passes through (x v ?/,), we have 

y l 7j 2 = 2a(x 1 + x 2 ) (2) 

Hence from (1) and (2) it is seen that the point (x 2 , y 2 ) 
always lies on the line 

yy l = 2a(x + x t ), 

which is therefore the required locus or the polar of (x v y t ) 
with respect to the parabola. 

When the point (x v y x ) is on the curve, the polar coincides 
with the tangent, 



148 



ANALYTICAL GEOMETRY 



When (x v y^) is outside, the polar is the chord of contact 
for that point. In general, the properties of the pole and 
polar proved in the chapter on the circle, will be found to 
be true for the conic sections. 

Note. — The polar of the focus (a, o) is x + a = [the directrix]. 
.-. tangents at the extremities of a focal chord meet on the directrix. 

Note on tangents: 

(I) To construct a tangent to a parabola at a given point P, take 




Fig. 110. 

MN = 2 a and draw PT _L to PN ; or take OT = OM and draw PT. In 
either way PT will be tangent at P. [Why?] 




Fig. m. 



(2) To draw a tangent to a parabola from an external point P. With 
P as center, and PF as radiue, cut the directrix in R. Draw RQ parallel 



THE PARABOLA 149 

to FX and cutting the parabola in Q. Draw PQ, which is the tangent 
required. For PF = PR, QF = QR, 

.*. PQ bisects angle FQR. .-., etc. 

(3) To find the locus of the intersection of two tangents which cut at 
a given angle \j/. 

Put x, y for x 1 , y v respectively, in the result of § 112, and we get the 
locus required ; viz., 

y 2 — 4 ax = (a + x) 2 tan 2 f, 

or 7/ 2 + (.r — a) 2 = {x + a) 2 sec 2 \f/ y 

which represents, as we shall see later, an hyperbola which has the same 
focus and directrix as the parabola, and whose eccentricity is equal to 
sec \f/. 

The equation may also be written thus, 

V?/ 2 + (x — a) 2 = (x + a) sec \p, 
which shows it more clearly. 

EXERCISES. 

1. In y 2 = 4 ax find equation of chord through vertex and which is 
bisected by the diameter y = a. Ans. a (y — x) = 0. 

2. The polars of all points on latus rectum meet the axis in the same 
point (— a, o). 

3. The line joining the focus to the pole of a chord bisects the angle 
subtended at the focus by the chord. 

4. Find the diameter of y 2 = 12 x, which bisects the chord 

z-3y-l = 0. 

5. Find the slope of the chords of a diameter of y 2 = 5 x which passes 
through (1, — 5). 

6. Show that the line y = m (x + a) -\ touches the parabola 

y 1 = 4 a (x + a) . 

7. _Pj and p 2 are ±s from the ends of a focal chord on the tangent at 
vertex. Prove p l • p 2 = a 2 . 

8. Find locus of the intersection of 

(1) Normals inclined at complementary angles to the axis. 

Ans. y 2 = a(x — a). 

(2) Tangents under same conditions. Ans. Latus rectum. 



150 ANALYTICAL GEOMETRY 

9. The length of a focal chord of a parabola is L. Show that the 
product of its segments = a • L [where y 2 = 4 ax]. 

10. Show that the locus of the mid-points of focal chords is 

y 2 = 2a(x — a). 

11. The locus of the mid-points of normal chords is 

y* - 2 ay 2 (x - 2 a) + 8 a 4 = 0. 

12. Two tangents are drawn to a parabola from a given point. Show 
that they are bisected by a third tangent parallel to the chord of contact 
of the first two. 

13. Taking the pole at the intersection of axis and directrix, find 
polar equation of parabola. 

14. A straight line is drawn from the intersection of the axis and 
directrix cutting the parabola, and a focal chord is drawn parallel to it. 
Prove, the rectangle of the segment of the focal chord is equal to the 
rectangle of the intercepts of the line made on the parabola. 

15. Find locus of center of a © which passes through a given point 
and touches a given line. [Take line and ± on it from point as axes.] 

Ans. A parabola. 

16. A point moves so that its distance from a fixed point always 
equals its distance from a fixed circle. Show that its locus is a conic 
having the fixed point as a focus. The fixed is called the " direction 
circle" of the conic. 

17. The tangents at the ends of a focal chord of any conic meet in the 
directrix. 

18. Given a diameter of a and a parallel chord. A line is drawn 
from the right end of diameter to mid-point of chord, and a line from 
right end of chord to mid-point of diameter. Find locus of their 
intersection. Ans. A parabola. 

19. F is a focus of a conic, P a point on the corresponding directrix. 
Prove PF is ± to the polar of P. 

20. Show that the polar of any point on the circle x 2 + y 2 = ax 
with respect to the circle, x 2 + if = 2 ax touches the parabola ^==4 ax. 



THE PARABOLA 



151 



116. Equation of the parabola referred to any diam- 
eter and the tangent at its extremities as axes. 




Fig. 112. 



SN is the given diameter, ST a tangent at its extremity S. 
We shall first find the equation referred to SN and SY' per- 
pendicular to it. 

m 
and putting this for y in y 1 = 4 ax, 

we get OM = — ■ . 

in 2 

.-. the point of contact S of any tangent 

is (— -, — ), which was also shown in f§ 1051. 
\ra J m / J 

or my 2 -\- 4 ay = 4 amx .... 



(i) 



is the equation of the parabola referred to SN and SY'; now 
let us retain SN as the #-axis and take SY"as the new y-axis. 



152 . ANALYTICAL GEOMETRY 

Let P be any point on the curve, (x, y) its co-ordinates re- 
ferred to SN and SY'j (x v y^ its co-ordinates referred to S.N 
and SY". 

Then x = SQ, x 1 = SR, 

y = PQ, y t = PR. 

Also, SQ = SR + PR cos 0, 

and PQ = PR sin 6, 

or, x = x t + y t cos 0, 

and y = y x sin 0. 

Hence, substituting these values in (1), and replacing m by 
tan 0, we finally obtain 



f = 



\a 



sin 2 
as the required equation. 

This equation may be simplified by other considerations ; 
thus, 



FS: 


= a -f OM = a + -^ 




(1 + m 2 ) 
m 2 




a sec 2 




tan 2 




a 




~ sin 2 ' 


Now let 


FS = a', 


then the equation 


becomes 




jr2 = 4 a'x- 



[referred to any diameter and the tangent at its extremity]. 

If = 90°, this equation becomes ?/ 2 = 4 aas, which is the 
equation referred to any diameter and a tangent JL to it, i.e., 
to the axis of the curve and the tangent at the vertex. 



THE PARABOL 



153 



117. Polar equation of the parabola referred to focus. 

P is any point (p, 0) on the curve. 
Now 
p = FP = PM 

= 2 a -f- p cos 0. . 
2a 



d' 


Y _ (p.Wp^* 




vT / 




M» ! 


N ( 
D 


,U u 



r i - cos e 

Discussion. — When 
d = 0, p = 00, 
. \ curve does not meet a: 

to right of 0. When = 

= 7T, cos 



0=2 

Summary. — As varies from to tt, p decreases from <x to a ; and 
as varies from -k to 27r, p increases from a to oo. 
Cos cannot be > + 1, . •. p is always positive. 

If the pole be taken at the vertex, the polar equation is p = — . 2 • 



3 


/7flr. 7/3. 


cos = 0, 


. •. p = 2 a [semi-latus rectum]. 


50= - 1, . 


■•. p = a [FO]. 


cos = 0, . 


•. p = 2 a. 


cos 0=1, . 


-, p = oo (as it should be). 



EXERCISES. 

1. Two tangents to y 2 = 4ax make ^/s ^, ^ 2 , with # = axis. 
Find locus of vertex if 

(a) sin \f/ x sin </>., = k (a constant) . 

a 2 , 2 o a 2 (1 - A; 2 ) 
^4ns. x 2 + ?/ 2 — 2 arr = p . 

(/3) tan ^j tan ^ 2 = fc. ;lns. a — /ex = 0. 

(7) ctn \f/ 1 ctn i/^ 2 = k. Ans. ka — y = 0. 

2. In y 2 = 4 a.r find the locus of 

(a) the mid-points of the ordinates. Ans. y 2 = ax. 

(j8) the mid-points of chords through the vertex. Ans. y 2 = 2 ax. 

3. PQ is a double ordinate of a parabola. The line from P to foot of 
directrix meets curve in P'. Show that P'Q passes through the focus. 



154 ANALYTICAL GEOMETRY 

4. Two parabolas have a common axis but different vertices. Show- 
that the part of a tangent to the inner, intercepted by the other, is 
bisected at the point of contact. 

5. Find the locus of the intersection of the tangents 

a 









y = 


= m 1 x + 


a 
m 1 


(«) 


m 1 




m 2 


= k. 




(0) 


1 


+ 


1 

m 2 


= k. 




(7) 


1 


- 


1 
m 2 


= k. 





+ 


m 2 




if 


Ans. 
Ans. 


kx - 
ka - 


- a = 0. 
-y = 0. 






A 


ns. 


y* = 


■■ 4 ax 


+ a 2 k 2 . 



6. Three tangents to a parabola, focus F, meet in A, B, C. Now, 
FA, FB, FC, meet BC, CA, AB, in A', B', C, respectively. Prove that 
the JLs from A, B, C, to the tangents drawn to the curve from A', B', 
and C, respectively are concurrent. 

7. n x , n 2 , n, , are the normals, and t x , t 2 , are the tangents from any 
point to the parabola y 2 = 4 ax. Prove, n x ■ n 2 • n 3 = a • t x ■ t 2 . 

8. In the parabola y 2 = 4 ax, tangents are drawn making with the 

tangent at vertex a A of constant area = k 2 . Prove that the locus of 

their intersection is, , , , . . . , . 

' x 1 (y 2 — 4 ax) =4 k\ 

9. If three normals are drawn from any point to a parabola, show 
that their feet and the vertex of the parabola are coney clic. 

10. Tangents are drawn at points whose ordinates are in the ratio 
r : 1. Find the locus of their intersection. 

Ans. ry 7 = ax (1 + r) 2 [where y 1 = 4 ax is the given curve]. 

11. Two tangents and their chord of contact form a A of constant 
area = k 2 . Find locus of their intersection. 

12. Find the locus of points from which two ± normals can be drawn 
to a parabola. 

Suggestion. — Show that normals at extremities of a focal chord are 
± to each other. Find locus of their intersection, etc. 

EXERCISES ON CHAPTER VII. 

1. Tangents are drawn to the parabola y 2 = 4 a^x. Find the locus of 
their poles with respect to the parabola y 2 = 4 a 2 x. Ans. a x y 2 = 4 a 2 2 x. 

2. The parabola referred to tangents at ends of latus rectum as axes is 
where ^^^j.^ 

k=2aV2. 



THE PARABOLA 155 

3. A circle cuts the parabola 

y 2 = 4 ax in 4 points, (x v y t ), (x 2 , y 2 ), ... etc. 

PrOVe y i + y 2+ y 3+ y i = . 

Suggestion. — Express that the four points arc concylic. Also each 
lies on y 2 = 4 ax. Eliminate x, etc.* 

4. A variable circle touches a given circle and a given line Show- 
that the locus of its center is the parabola, focus at center of given 0, 
directrix parallel to the giveu line, and at a distance r from it, where r 
is radius of the given circle. 

5. A secant revolves about a fixed point on the axis of a parabola, 
and normals to the curve are drawn at the points of intersection with the 
secant. Find the locus of their intersection. 

6. Construct a parabola given the directrix and two of its points. 

7. Find the locus of the vertex of a parabola which has a given focus 
and touches a given straight line. 

8. Find the locus of its focus if it has a given vertex and touches a 
given straight line. 

9. Find the locus of a focus of a parabola which touches two given 
straight lines, one in a fixed point and the other in a variable point. 

10. Find the locus of the foci of parabolas which have a common 
chord and a common tangent parallel to this chord. 

11. The sides of a quadrilateral inscribed in a parabola are inclined 
to the axis at angles a, /3, y, 5, respectively. Prove : 

ctn a + ctn y = ctn /3 + ctn 5. 

12. Find the locus of the point of intersection of two parabolas which 
have a given focus, which touch a given line, and which intersect at a 
given angle 0. 

13. Show that the line x cos a + y sin a -\ == touches the para- 

y cos a r 

bola y 2 = 4a (x + a). 

Suggestion. — Compare x cos a 4- y sin a = p with the equation of a 
tangent to the parabola at (x u ?/,). What is this equation ? Find for 
what value of p the line will be a tangent, etc. 



* See Ex. 97, page 269. 



156 ANALYTICAL GEOMETRY 

14. If K is niid-point of chord PQ of a parabola, show that the polar 
of R is parallel to PQ. 

15. P, Q, R, are three points on the parabola y 2 = 4 ax whose abscissae 
are in geometrical progression. Show that the tangents at P and R 
meet on the ordinate of Q. 

16. The locus of the mid-point of that part of a variable tangent to 
a parabola which is intercepted between two fixed tangents is a straight 
line. 

17. The parabola y 2 = 4 ax slides between two rectangular tangents. 

Show that its focus traces the curve x 2 y 2 = a 2 (x 2 + if) . 

a x 

Suggestion. — Let the tangents be y = mx -\ and w = — — — am. 

Find the distances from focus (a, o) to these tangents. Call these x and 
y, and eliminate ra, obtaining the above equation of the locus referred to 
the tangents as axes. Show also that the vertex traces the curve 
x% y% + x s y* = a 2 . 

18. The equation of a chord whose middle point is (x v yj is (y — y^) 
y 1 = 2a(x- x t ). 

19. The locus of the mid-points of normals is y 2 = a (x — a). 

20. The locus of the mid-points of chords through (x v y x ) is the 
parabola, y (y — y^) = 2 a (x — x x ). 

21. The length of the chord of contact of tangents from (h, k) is 

1 ,. 

-v / (A; 2 +4a 2 ) (A; 2 -4a/i) 

22. A circle passes through the focus of a parabola and cuts the curve 
at angles whose sum is constant. Find the locus of its center. 

Ans. A straight line. 

23. TP, TQ, are tangents to a parabola ; the circle through T, P, Q, 
meets the parabola again in P' and Q'; and the tangents at P' and Q', 
meet in T\ Prove TT' passes through the focus. 

24. Find the equations of the tangent and normal at the upper end of 
the latus rectum of the parabola y 2 = 4 ax. Ans. y — x = a. 

y + x =3a. 

25. Prove that the product of the latera-recta of two parabolas which 
can be described through four concyclic points is \ (d 2 — d 2 2 ) sin <f> 
where d 1 and d 2 are the diagonals of the quadrilateral and <j> the angle 
between them. 



THE PARABOLA 157 

26. The lines joining the origin (vertex) to the points of contact of 
tangents to a parabola from (x v y t ) are represented by the equation 

x<y 2 = 2x {y t y - 2 ax). 

27. A ± Ap from vertex to a tangent at P meets the curve in q. 
P: ove : Ap • Aq = 4 a 2 . 

28. The third diagonal of a quadrilateral inscribed in a circle, and 
circa inscribed about a parabola, passes through the focus of the parabola- 

29. T and T' are two external points. TP, TQ, T'P', T'Q', are tan- 
gents to the parabola. If TT' is bisected by the parabola, show that 
the six points T, P, Q, T', P', Q', lie on another parabola. 

30. Tangents at P, P', meet in T. Prove : 

TP 2 FP 

■ = = (where F is focus), 

TP /2 FP' 

Also FT 2 = FP • FP'. 

31. A chord PQ [of a parabola], normal at P, meets the axis in M. 
If y is the ordinate of P, prove 

PM 4 
AFPQ = f^-. 
4 ay 

32. Given the base =2 a (of a A), altitude = -, find the locus of its 
orthocenter. 

Suggestion. — Take mid-point of base as origin. 

Ans. x 2 + by — a 2 = [a parabola]. 

33. FL is the J_ from focus on tangent at P. A circle is circum- 
scribed about the A FLP. Show that the locus of its center is the par- 
abola 



CHAPTER VIII 



THE ELLIPSE 

118. Definition. — If P is a point on the curve, we have 
FP = e • PM, where F is the focus, PM = _L on DD' [the 
directrix] and e < 1. 

Y 




B 

Fig. 114. 

119. To find the equation of the ellipse. — Draw FN J_ to 
the directrix. Now divide FN, internally at A, and exter- 
nally at A', in the given ratio e : 1. 

.'.by definition A and A' are points on the curve, since 

FA. = e • AN 

and FA' = « . A'N 

Now bisect A A' in 0, calling the distance AA' 2 a. 
Add equations (1) and (2), we get, 

2 OA = e • 2 ON, 

2a = 2e ■ ON, 

158 



(1) 
(2) 



or 






(3) 



THE ELLIPSE 159 

and subtract (1) from (2), we get, 

2 OF = e • 2 OA, 

.\OF = ae (4) 

Now, through draw BB' _L to AA'. Take OA' and OB' as 
axes. The co-ordinates of F are (— ae, o). 
Let P be any point (x, y) of the curve. 

Then FP 2 = (x + ae) 2 + y 2 , 



a 



also PM = ON + OQ = x + - 

e 

Now, FP 2 = e 2 • PM 2 

or, (x + ae) 2 + y 2 = e 2 ( a; + - ) • 

... x * (1 _ 6 2) + y/ 2 = a 2 (1 - e 2 ) .... (5) 
The intercepts of this on the ?/-axis are both real and = 

± a VIT^ 2 - OB r . 
Let OB' = ft, • ' . h 2 = a 2 (1 - e 2 ). 

Dividing (5) by its right member [ft 2 ], we get, 

S+? =i ( 6 > 

which is the required equation of the ellipse. 

Discussion. — Equation (5) reveals the following facts about the 
curve : 

(1) It is symmetrical with respect to the z-axis, i e., the line through 
the focus and ± to the directrix. This line [A A'] is .-. called the major 
or principal axis. 

(2) It is symmetrical with respect to the ?/-axis, i.e., to [BB'J a line 
± to AA' at its middle point 0. This line is . \ called the minor axis. 

(3) The greatest abscissas are -j- «, ordinates -[- b, and no real values 
for x and y beyond these. The ellipse is .-. a closed curve of one 
branch lying entirely on the same side of the directrix as the focus. 

(4) It is symmetrical with respect to 0, which is .-. called the center. 



160 ANALYTICAL GEOMETRY 

(5) The ellipse has a second focus at F' and a corresponding direc- 
trix DD'. F' is point (ae, o). The equations of the directrices are 

s±| = 0. 

(6) The latus rectum [double ordinate through focus] is equal to 2 b 

2 b 2 

y/l — e = [found by putting x = ae in equation (5) or (6), then 

a 

solving for ?/, and multiplying by 2]. 

Note. — If the center is at the point (h, k), the equation becomes 
(* - h? (y - fc) 2 _ 

a 2 + 6 { } 

Query. — Which ellipses are represented by these equations ? 

S+5- 1 ( 8 > 

Or- A) 2 (y-fr) 2 (9) 

6 2 + a? ~ 
Note 1. — The distance FF' is generally denoted by 2 c. 
Note 2. — A and A' are called the vertices of the ellipse. 

120. Every equation of the form Ax 2 + By 2 + 2 Gz + 
2 Fy + C = represents an ellipse. — We shall illustrate this 
by a special example ; viz., 

4 X 2 _j_ y 2 _ 3 £ _j_ 2y -j- 1= 0. 

This may be written, 

4cc 2 -8x + ?/ 2 + 2?/ + l = 0, 

4(z 2 - 2x + 1)+ (> 2 + 2y + 1) = 4, 

or 4 (,x - l) 2 + (y + l) 2 = 4, 

or <* ~ !) 2 . (y ± !) 2 _ i 

(If + (2) 2 

Comparing this with the type form (7) or (9), we see that it 
represents an ellipse whose center is at point (1,-1) and. 
whose semi-axes are a = 2, b = 1. The ellipse has its major 
axis parallel to the ?/-axis. 



THE ELLIPSE 161 

The other facts about this curve are readily found; viz., 
the foci are the points (1, — 1 -f- VB), (1, — 1, — V3) ; the 
vertices are the points (1, 1), (1,-3); the latus rectum is 1. 

The general equation at the beginning of this article may 
also be shown to represent an ellipse by completing the 
squares, etc. 

121. Internal and external points. — In a manner similar 
to that used in the parabola, we may show that a point (x, y) 
is without, on, or within the ellipse 

x2 ,V 2 _-i 
a 2 + b 2 ~ lt 

according as — + ^ — 1 > , =, or < 0. 

122. The squares of any two ordinates are to each 
other as the products of the segments into which they 
divide the major axis. — Let (x 1} y^) and (x 2 , y 2 ) be any two 
points on the ellipse. Then, 

b 2 b 2 

V? = ~2 (« 2 - *i 2 )> V? = - 2 <> 2 - ^ 2 2 ). 

•*• V\ ' V? '• '• — ffi) (« + «i) : (a ~ *" 2 ) + x z)- •'•> etc - 

123. The Latus Rectum is a Third Proportional to the 

2 b 2 4 b 2 
Major and Minor Axes. — The latus rectum = — = - — 
J a 2a 

[Let c ioe the latus rectum.] 

.-. 2a :2b :: 2 b : I. .'., etc. 

124. The sum of the focal distances of any point of 
an ellipse is constant and equal to 2 a. — In the figure of 

§ 116, FP = e . PM = e . QN = e(ON + OQ) = e (- + xY 

.-. FP = a + m. 



162 



ANALYTICAL GEOMETRY 



Also, FP = e.PM'=e.QF=e. (ON'- OQ). 

.-. F'P = a - ex. 
.-. PF + PF'=2a. 

Hence, the ellipse is the locus of a point which moves so 
that the sum of its distances from two fixed points [the foci] 
is constant and equal to 2 a. 

125. Construction of the ellipse when foci and the 
constant sum 2 a are given. — This is based on § 123. 

(1) By points. 




Bisect FF' in 0, and take OA = 0A'= a. Then A and A' 
are points on the curve. [Why ?] Now between F and F' 
take any point K. With F as center and AK as radius, 
describe arcs ; with F' as center and A'K as radius, describe 
arcs intersecting the former in P and P'. By interchanging 
radii, Q and Q' are found. These all lie on the curve because 
the sum of the focal distances of each is 2 a. After finding 
enough points we join them by a smooth curve. 

(2) Mechanically. Fix pins at F and F' and join them by 
a string of length 2 a. A pencil point P moved so as to keep 
the string stretched will trace the ellipse ; for in all positions 
of P, we have PF + PF'= 2 a. 



THE ELLIPSE 163 



EXERCISES. 



1. A line AB of given length moves with its ends in two rectangular 
axes OX, OY. Find the locus of a fixed point P on the line. 

Put PB = a, PA = b, /_ OAB = 6. 

Draw PN ± OX. Then the co-ordinates of P are (x, y) ON, PN. 

Then sin 6 = =— = ■{ , cos = - • 

PA b a 



Squaring and adding, 



^ hi 



Note. This principle is utilized in an instrument for describing the 
ellipse mechanically. 

2. If OP and OQ are two semi-diameters ± to each other, prove, 

1 J_ 1 1 
OP 2 OQ 2 _ « 2 & ' 

3. Find equation of ellipse, focus (1, 2), directrix 2x — y — 1 = 0, 
eccentricity = \. 

4. Is the point (2, 3j inside or outside the ellipse x 2 + 4 y 2 = 16 ? 

5. Find eccentricity, length of latus rectum, co-ordinates of foci in 
the ellipses. 

(1) 25 x 2 + 9 if = 225. 

(2) 9.x 2 + 4 if = 36. Ans. J VS, 2§, (0, J-V5). 

(3) 4(.r-2)2 + 9(^/-3) 2 = l. 

(4) 2.r 2 -i- if =3.r. Ans. — = = eccentricity; foci, (1,^1). 

V2 

(5) x 2 + 9y 2 = Sl. 

6. P, Q, K, are three points on the ellipse whose abscissas are in 
arithmetical progression. Show that OP, OQ, OR, are also in arith- 
metical progression. 

7. P is any point on the ellipse, and PB, PB', meet A A" in p, q. 
Prove Op ■ Oq = a 2 . 

8. Equation of the ellipse with A as origin is y 2 = — (2 ax — .r 2 ). 

9. Show by Ex. 8 that locns of mid-points of chords of the ellipse, 
passing through A, is another ellipse. 



164 ANALYTICAL GEOMETRY 

10. Given base of A ABC and tan J A • tan J B, find locns of vertex C. 
tan | A • tan | B= — . \ S [sum of sides] is given. 

Ans. Ellipse, foci A and B [ends of base]. 

11. Find also the locus of center of its inscribed 0. 

Ans. Ellipse, major axis = AB. 

126. Note 1. — We may show that the circle is a special case of the 

x 2 v 2 
ellipse, thus: If, in the equation-^ + j- 2 = 1, a remain constant, while 6 

increases, then c decreases [for c =Va 2 — 6 2 ], 



■. e decreases, for 



[«-«,.-..-£] 



. \ the foci approach the center, and the ellipse approaches a circle. 

If 6 = a, c = o, . •• e = o, ••• foci coincide with the center, and the 
equation of the ellipse becomes x 2 + y 2 = a 2 , a circle, radius a. 

If o decreases and approaches o, c approaches a, and finally when 
6=o, c = a, e = 1. Hence the ellipse becomes thinner gradually, and 
finally coincides with the z-axis, for its equation becomes y =■ o. 

Note 2. — To find the polar equation of the ellipse, with center as 
pole, we put y = p sin 6, x = p cos 0, and obtain 
p 2 cos 2 p 2 sin 2 
a 2 + 6 2 = ' 



or, - 2 = — u— + 

a- b 



1 cos 2 sin 2 
? 



a 2 b 2 



r a 2 sin 2 + b 2 cos 2 
a 2 b 2 
~ b 2 + (a 2 - 6 2 ) sin 2 ' 
This denominator is least [?'. e., p is greatest] when = o ; .-. p = a. 
.-. the ellipse is a closed curve, which was also shown before. 

127. The tangent at (x 1 , i/ t ). — The equation of any chord is 
Also J+ |i = l (2) 

S+S =i ( 3 ) 



THE ELLIPSE 165 

'.by subtraction 

i 
~a 2 



(x t -x 2 ) (x x + x 2 ) (y l - y 2 ) (y t + y 2 ) _ ft 

,2 T" /,2 — U 



.". (1) becomes, 



y — yi _ __ £ 2 gj + g 



*- ...... (4) 

x — a^ a 2 y 1 + ?/ 2 



Now put x 2 = x 1} y 2 =y x , then (4) becomes 
V - Vx _ __V x i 



(5) 

x — x t a- y 1 



or, a 2 yy x -f- b 2 xx x = a 2 y 2 + b 2 x x 



01 



X Jh 4. Mi = r^L 4- ?l1 = I 
a 2 "*" £ 2 "" \_a 2 ^ 6 2 J 

' * a 2 "^ 6 2 



The normal is (y — y t ) = ^-^ (x — x^ 



is the required equation. 

- *i>i = 

b 2 x.^ 

The subtangent and subnormal are found by putting y = 

in equations of the tangent and normal; viz., 

jv 2 ^2 

subtangent = — 

x i 

subnormal = — — -x, • 

a- 

Note. — That the subtangent is a function of the abscissa and a. 

128. Equation of tangent in terms of slope. — The abscis- 
sas of the points of intersection of the line 

y = mx -\- c ., (1) 

with 5+S =i ( 2 ) 

are determined by the equation 

x% . ( mx + ^ _ i rT i 

a« + P ~ * (d) 



166 ANALYTICAL GEOMETRY 

The roots are equal ; i.e., (1) is a tangent to (2) if in the 
equation 

(a 2 m 2 + b 2 ) x 2 + 2 me a 2 x + a 2 (e 2 - b 2 ) = 0, 
which is the equivalent of (3), we have 

(a 2 m 2 + b 2 ) ■ a 2 (c 2 - b 2 ) = m 2 c 2 a 4 , 
or, c 2 = a 2 m 2 + b 2 . 



.-. c = ± \la 2 m 2 + b 2 . 

Hence, there may be two tangents of given slope m to the 
ellipse ; viz., y = mx ± ^ ^ + ^ ^ 

The point of contact of this tangent may be easily found by 
comparing the equations 

y = mx ± Vm 2 a 2 -f- b 2 

and *4 + *& = 1, 

a 2 b 2 

as in the case of the parabola, and finding x x , y x in terms oi 
a, b, and m. 

129. Tangents from a given point. — If (4) passes through 
(x 1} Vl ) we have ^ = ^ + y m 2 a« + fc 2 , 

whence m = ^ ± V ^^ + ^ ~ a * I . 



.'. There are two real, two coincident, or no tangents, 
according as b 2 x 2 -j- a 2 y 2 — a 2 b 2 > , = , or < 0; i.e., accord- 
ing as (x^y^) is without, on, or within the ellipse. 

EXERCISES. 

1. A A is circumscribed about an ellipse. Show that the products of 
the alternate segments of the sides made by the points of contact are 
equal. 

2. In a given rhombus to inscribe an ellipse of given eccentricity. 



THE ELLIPSE 167 

3. The tangent and normal to the ellipse at the end of the latus 
rectum [first quadrant] are 

y + ex — a = 0, cy — x + ae 3 = 0. 

4. For what value of k will the line x + y = k touch the ellipse 



X - + t = 1 ? 
a 2 + 6 2 

5. A square is inscribed in an ellipse. Find length of its side. 

2ab 
Ans. 



Va 2 + b 2 

6. Tangents to the ellipse from (h, k) make an angle <j>. 
Prove, \ h 2 + fc 2 - a 2 -b 2 \ tan 0=2 Vb 2 h 2 + a 2 k 2 - a 2 b 2 . 

7. P is any point on the ellipse. Show that the locus of the center of 
the circle inscribed in A FPF' [F, F' are foci] is the ellipse 

(1 - e) x 2 + (1 + e) y 2 = e 2 a 2 (1 - e). 

8. The locus of the mid-point of a normal is the ellipse 

4 b 2 x 2 + 4 a 2 y 2 (1 + e 2 ) 2 = a 2 b 2 (1 + e 2 ) 2 . 

Note. — Normal here means length of normal intercepted by major 
axis. 

9. The locus of the mid-points of chords of the ellipse which touch 
the concentric circle x 2 + y 2 = c 2 is the curve 



a- b 2 ) (a 4 ' ¥ 

10. A A inscribed in an ellipse has a fixed centroid. Show that its 
orthocenter traces an ellipse. 

130. The auxiliary circle. The principle of the elliptic 
compass. 

The circle described with the center of the ellipse as a 
center and a radius OA is called the major auxiliary circle. 
That with a radius OB, the minor auxiliary circle. We shall 
consider only the former for the present. 

Let P (x, y) be any point on the ellipse, whose ordinate PN 
meets the auxiliary circle in Q. Now from the equation of 
the ellipse we get 



168 



ANALYTICAL GEOMETRY 
Y 



—x 




or. 



PN = - VoQ 2 - ON 2 = -. QK 



a a 

.-. FN : QN* : : ft : a. 

Hence, the ordinates of the ellipse and auxiliary circle at 
corresponding points are in the constant ratio of b : a. 

Now draw PCM II to OQ and cutting the axes in C and M. 

Then PM = OQ = a, 

and by similar As, PC : OQ : : PN : QN : : b : a. 

.-. PC = b. Hence, if PM be a ruler having pins at 
C and M which may move along the ruler at will, and if 
AA', BB', are grooves in which these pins run, a pencil point 
at P will trace an ellipse whose semi-axes are PM and PC, or 
a and b. Such an instrument is called an elliptic compass.* 

131. Eccentric angle. Chord joining two points. — The 
angle QOA' [Fig. 116, § 130] is called the eccentric angle of 
the point P. The co-ordinates of P may be written x — a cos 



* See Fig. 173, end of book. 



THE ELLIPSE 



169 



^ y = b sin \p, where \p is its eccentric angle ; for these values 
evidently satisfy the equation 



_ + v_ 

a 2_h b 2 



1. 



Again, two points whose eccentric angles are ^ and if/ 
respectively, may be written (a cos i//, b sin i//), and (a cos i//, 
b sin if/') ; and the chord joining them is evidently 

x 2/1 

a cos \p b sin \p 1 
a cos »// b sin i// 1 



= 0, 



or bx (sin ^ — sin i//) — az/ (cos i// — cos i//) = ab sin (^ — i//), 



or 2 6# sin 



(*?■) 



cos 



= 2 aS sin 



( ±±£ 



Divide by 2 ab sin 



<A 



■ 2 ay sin 

1 cos ' 



* + v 



f 



)+m 



and we get as a final result, 



cos 



[* + £] 



V ■ 
+ 7 sin 
b 



[> + *H 


= cos 


[> - f 1 


2 




2 



which is the equation of the chord joining the two given 
points. 

Note. — We shall find that by using the eccentric angle, we obtain 
symmetrical results which shorten the work very much. 

132. Equations of tangent and normal at the point whose 
eccentric angle is t//. — If the chord in § 131 revolve about 
one of the two points given, it will become a tangent at that 
point, and ultimately i// = \p. Hence, to find the equation of 
the tangent, we put i// = \p in the result of § 131, and obtain 



x ?/ 

- cos ib + V sin \b = 1 

a T b 



(i) 



170 ANALYTICAL GEOMETRY 

The normal is _L to it at the point {a cos if/, b sin if/) ; its 
equation is .'. found to be 

^L_j2L =a *_ i2 (2) 

COS if/ Sill if/ w 

The tangent at the vertex (a, o) is found thus : the eccentric 
angle at that point is ^ = 0, .*. (1) becomes 

x 

- cos ib = 1, or, x = a. 
a 

The normal, 

ax sin if/ — by cos if/ = sin if/ cos if/ (a 2 — b 2 ), 

becomes, — by cos if/ — 0, by = 0, .-. y = 0. 

These results are also obtained by putting (a, 6) for (a^, 2/ x ) 

in the equation — ± + ^i = 1 * 

a 2 & 2 



133. Locus of the foot of a _L from the focus to a tan- 
gent. — Any tangent is 

y = mx -\- \/m 2 a 2 -j- b 2 (1) 

The _L on it from the focus (ae, o) is 

x -h my = ae . . . . . (2) 

Squaring and adding (1) and (2), we get, 

(1 + m 2 ) (x 2 + y 2 ) = a 2 m 2 + b 2 + a 2 e 2 
= a 2 (1 + m 2 ). 
.-. sc 2 + y 2 = a 2 is the required locus [the auxiliary circle]. 

EXERCISES ON ECCENTRIC ANGLE. 

1. The locus of the intersection of tangents to an ellipse at points 
whose eccentric angles differ by a constant 2 ^ is the ellipse 

3 + £-■**. 



* Equation (1) above is also obtained from this equation by putting 
x x = a cos \j/, y 1 = b sin \j/. 



THE ELLIPSE 171 



2. The point of intersection of the tangents 



x y . ., \ 

-COS a + t sin a = 1, 
a 6 



cos /3 + | sin 
o 



at points whose 
eccentric angles 
are a and /3, 



is 




8,n (-T 



C-i- e )J 



3. P is a point on the ellipse, eccentric angle = a. Show that the 

equations of A'P and AP are 

x a y . a a 

- cos - -f i sin - = cos -, 

y a x . a .a 

£ cos sin - = sm -. 

b 2 a 2 2 

Suggestion. — Eccentric /_ of K! is 0, of A is tt. 

4. a and /3 are the eccentric angles of the extremities of a focal chord. 

Prove tan - • tan - = • 

2 2 e + 1 

5. yp x and ^ 2 are the eccentric angles of the points of contact of 
tangents from (A, k). 

6. A chord is drawn joining the points whose eccentric angles are a 
and /3. Show that its length is 2 5 sin ( — — -^-J where 5 is the parallel 
semi-diameter. 

7. The vertices of a A are three points on the ellipse whose eccentric 
■angles are a, /3, y. Show that its area is 

8 The area of the A formed by tangents at the points whose eccen- 
tric angles are a, /3, 7, is 

ab ■ tan \ (a — /3) tan \ (/3 — 7) tan £ (7 — a). 

9. Two chords of the ellipse meet the major-axis at points equidistant 

from the center. The eccentric angles of their extremities are a, /3, 7, 5. 

t. a fl 7 5 , 

Prove tan - tan g tan | tan - = 1. 



172 ANALYTICAL GEOMETRY 

10. M is foot of the ± from center on a tangent at P. From M a 
tangent is drawn, touching the ellipse in Q. The eccentric angles of P 
and Q are \j/ and <f> respectively. 

& 2 _ tan \p 
Prove a 1 



tan 



m 



11. P and P' are two points on the ellipse, whose eccentric angles are 
0, and 2 . The circle describes on PP' as a diameter meets the curve 
again in Q and Q'. Show that the equation of QQ' is 

134. Locus of the intersection of two tangents _L to each 
other. — Let one of the tangents be 

y — mx = VfflV -h b 2 (1) 

Then the other is y — m x x = \Jm 2 a 2 -\- b 2 , 

but m. = 

1 m 

.*. its equation is, my -f- x = \]m 2 b 2 + a 2 .... (2) 

Squaring and adding (1) and (2), we obtain 

(1 -f m 2 ) x 2 + (1 + m 2 ) y 2 = (1 + m 2 ) (a 2 + b 2 ). 

.-. x 2 + y 2 = a 2 + b 2 

is the required locus. It is called the director circle. An- 
other method : If the line x cos a + V sin a = p is a tangent 
to the ellipse, its coefficients and those of the equation 

— l -JUL" - 1 

a 2 "•" 6 2 

must be proportional. 

a?, ?/i 1 



* a 2 cos a Z> 2 sin a p ' 
a cos a 6 sin a jj a cos a -f- # sin a " 



THE ELLIPSE 



173 



But since 



&MH 



i, 



.:p< 



COS 2 a + 6 2 sin 2 a. 



,-. p = Va 2 cos 2 a + b 2 sin 2 a. 
.-. the line x cos a + y sin a = \la 2 cos 2 a + b 2 sin 2 a 
is a tangent to the ellipse. Now put - ~\- a for a, and we obtain, 



— # sin a -f- y cos a = Va. 2 sin 2 a -\- b 2 COS 2 a 
for the equation of the tangent _L to the first. 

Squaring and adding both, a is eliminated, and we obtain, 
x 2 -+- if — a 2 + b 2 , as before. 

135. Properties of the ellipse. — (1) Tangents drawn at 
corresponding points, to any number of ellipses [circle in- 




Fig. 717. 



eluded] which have a common major axis, meet on that axis 
produced. For the sub-tangent 



is constant for all. 



l-'-^rl 



174 ANALYTICAL GEOMETRY 

(2) The normal at any point bisects the angle between the 
focal radii of that point. 

XX 1/11 

The tangent at P is -± + ^ = 1 ; 

and putting y = 0, 

we get xx x = a 2 , i.e., OM • OT = a 2 . 

Similarly, yy x = h\ i.e., PM • OT x = b 2 ; 

and putting y = in the equation of the normal, we get 
x = xJl - -\ = e 2 x, , i.e. , ON = e 2 • OM. 

.-. F'N = ae + e 2 x 1 = e • FT, 
and FN = ae - e 2 x i = e . FP. 

.♦. F'N : FN : : F'P : FP. 

.-. PN bisects ZF'PF [by Plane Geometry]. 

.-. , etc. 

(3) If OR is the _L from origin on the tangent, then 
PN • OR = b 2 . 

The distance from P (x v y { ) to N (e 2 x v 0) is 



Vfo - e\) 2 + y 2 = \Jx 2 ^ + y 2 

and the _L from origin to the tangent = OR is equal to 

1 



J 



Vx 



1 

A ' 



a- ¥ 
.-. PN. OR = 6 2 . .-., etc. 

(4) If F'L' and FL are _l_s on the tangent from the foci, 
then, F'L' . FL = b 2 [proof by student]. 



THE ELLIPSE 



175 



(5) 1/ and L lie on the auxiliary circle. Let FL meet F'P 
produced in Q. Then As PLF and PLQ are equal. [Why ?] 

.-. LF = LQ, PF = PQ. .-. F'Q = 2 a. 

. . FQ = 2 FL, and F'F = 2 OF. 

.-. OL is parallel to F'Q, and = i F'Q = a. 

Similarly, OL' = a. 

.-. 1/ and L are at a distance a from the center. 

.-. L' and L lie on the auxiliary circle. 

Note. — For variety, the student should prove the various exercises 
both analytically and geometrically when possible. The analytical proof 
will generally be found more elegant and more expeditious. 

(6) The area of an ellipse 
is 7r ab. 

Describe the auxiliary cir- 
cle. 

Then, 
rectangle MPKN MP _ b 

a 



rectangle MQSN MQ 






Y~f~ 


> "'*>k 


/pf^r 


L s 


f s 


>v \ 


1 / 


X \ 


If 


\\ 



UNO 

Fig. 118. 



Hence, the sums of any number of rectangles with equal 
bases will be to each other as b : a. 

.-. the elliptic and circular quadrants have this ratio, by 
the theory of limits. 

ellipse b 
a 



circle 
But circle = it a?. ,\ ellipse [area] 



ab. 



136. Equation of diameter of ellipse. 

Let any chord of a parallel system be y = mx -f c, and 
( x v Vi)) ( x 2» 2/2)) its extremities. 



176 ANALYTICAL GEOMETRY 

Then the co-ordinates of its mid-point are found from these 
equations, 

— ' Ju — — Ju* ~j~ tZ/2 ) 

ly^Vx + y*' 
Also, ,. m = yi^li = -^. x ±±^. 

x x - X 2 a 2 y^ -f y 2 

b 2 x 
.*. w = =-• 

a 222 

is the equation of a diameter, a straight line through the 
center. 

137. Conjugate diameters. — If in the preceding article 
we write the equation of the diameter in the form 

y = m 1 x, then m x = - ^ ■ 

b 2 
1 a 2 

This relation is evidently the condition that the diameter 
y = m x x shall bisect all chords parallel to the diameter y = mx. 
It is also the condition that the latter shall bisect the chords 
parallel to the former. 

Hence, if one diameter bisects all chords parallel to an- 
other, the second bisects the chords parallel to the first. Two 
such diameters are called conjugate diameters. 

The condition that two diameters be conjugate is again, 

b 2 
1 a 2 

This condition shows that the slopes of two conjugate 
diameters have opposite signs, .-. two conjugate diameters 
of an ellipse lie on opposite sides of the minor axis. 



THE ELLIPSE 177 

As an exercise, let the student prove, both geometrically 
and analytically, that the tangents at the extremities of any 
diameter are parallel to the chords of that diameter. Also 
let the student construct a tangent and normal to an ellipse at 
a given point. 

EXERCISES. 

1. Two focal chords are drawn at right angles in an ellipse. If r v 
r 2 , are their lengths, prove 

11 . , a (2 - e 2 ) 

— — = a constant = — ^— — — - . 
r x r 2 2 b 2 

2. A tangent at the end of the latus rectum of a certain ellipse 
passes through a point of trisection of the minor axis. Prove that the 
eccentricity of the ellipse is determined by the equation 

9 e 4 + e 2 - 1 = 0. 

3. An ellipse slides between two rectangular axes. Show that the 
locus of its center is the circle 

2? + 7/2 = a 2 + h 2; 

Take two ± tangents [in slope form] as axes. Put their distances 
from center (o, 6) equal to x and y respectively. Eliminate m, etc. 

4. The two tangents from (h, k) to the ellipse are represented by 
the equation 

Vh 2 k 2 1 ~| f.r 2 y 2 1 ~| \~hx | ky 
la 2+ F 2 



-] B-6-0-GF+*- 1 ! 



5. The locus of the foot of al from the center of an ellipse to a 

tangent is the curve 

(. r 2 + y*y = a 2 x 2 + bhf. 

6. A tangent meets the director circle [x 2 + y 7 = a 2 + b 2 ] in M and 
N. Prove OM and ON are conjugate diameters of the ellipse [where 
is center]. 

7. Each of the two tangents drawn to an ellipse from a point on the 
directrix subtends a right angle at the focus. 

8. The circle on any focal distance as a diameter touches the major 
auxiliary circle. 



178 ANALYTICAL GEOMETRY 

9. From the center of an ellipse two radii vectors are drawn ± to 
each other, and tangents are drawn to the ellipse at their extremities. 
The locus of the intersection of these tangents is the ellipse 

t a- £ = I + I 
a 4 + 6 4 a 2 "*" 6 2 " 

10. Locus of the intersection of tangents at the ends of two conjugate 
diameters is the ellipse 

£ + £ = 2 . 
a 2 ^ 6 2 

11. Find a point on the ellipse at which the tangent makes equal 

angles with the axes. . ( a 2 b 2 \ 

Ans. - 

\Va 2 + 6 2 \Za 2 + b 2 ) 

Also, so that the tangent makes intercepts proportional to the axes. 



Ans. 



(—• — )■ 

\\/2 V'2/ 



12. The tangents to an ellipse from any external point make equal 
angles with the lines drawn to the foci from that point. 

13. A parabola, described with any point on an ellipse as focus, and 
the tangent at the corresponding point to the major auxiliary circle as a 
directrix, passes through the foci of the ellipse. 

14. The ± from focus on a tangent, and the line joining the center to 
the point of contact, meet on the corresponding directrix. 

15. P, Q, R, are three points on an ellipse; p, q, r, are the three cor- 
responding points on the auxiliary circle [major]. 

Prove A PQR : A pqr : : b : a. 

16. Find the locus of a point such that tangents being drawn from it 
to an ellipse, the area of the A formed by these with major axis = A 
which they form with minor axis. 

.4ns. The equi-con jugate diameters. 

17. Any two conjugate diameters subtend angles 1? 2 , at a fixed 
point on the ellipse. 

Prove cot 2 d 1 + cot 2 6 2 = a constant. 

18. PQ is a focal chord, and P is the intersection of the tangent at P 
and the normal at Q. Show that QR is bisected by the minor axis. 

19. P is any point (x, y) on the ellipse. Show that the ctn of angle 
APA' varies as y. 



THE ELLIPSE 179 

20. A line through parallel to the tangent at P meets the focal 
radii of P in M and N. 

Prove PM = PN = a. 

21. P and Q are two points on the ellipse the sum of whose eccentric 
/y> = 2 a [a constant] . Find the locus of intersection of tangents at P 
and Q. Ans. [Straight line] ay — bx tan a = 0. 

22. The normal at P meets the axes in L and J/. 

Prove PL • PL' = PF • PF' [where F, F' are the foci]. 

23. P is any point on the ellipse. Show that the locus of the inter- 
section of AP with the _L to A'P through A' is the straight line. 

_^ a{a 2 + b 2 ) 
X (a 2 - b 2 ) 
24.. OP, OD, are semi-conjugate diameters. 
Prove FP • FT = OD 2 . 

Show also that the locus of the mid-point of PD is 
x 2 y 2 1 
a~ 2 + ¥ = 2 

25. The locus of the mid-points of chords through (h, k) [a fixed 
point] is the ellipse 

x 2 y 2 _ hx ky 
a 1 + 'b 2 = H? + t> ' 

26. The normal at P meets the major axis in M and the diameter 
conjugate to OP, in N. 

Prove PM ■ PN = b 2 . 

27. The equation of the chord whose mid-point is (h, k) is 

^( x -h)+^(y-k) = 0. 

28. P and J) are the ends of conjugate diameters. Prove that the sum 
of the squares of the ±s from P and 1) on a fixed diameter is constant. 

29. N is any point on the auxiliary circle [major]. A'N, AN, meet 
the ellipse in JVT and M. 

AN A'N a 2 + b 2 
AM A'M' b 2 

30. Show that the normals at the ends of a focal chord intersect on a 
parallel to the major axis through the middle point of the chord. 

31. From any point P on the ellipse ±s are dropped on the equi- 
conjugate diameters. Show that the normal at P bisects the line joining 
their feet. 



180 



ANALYTICAL GEOMETRY 



32. In an ellipse chords are drawn of a constant length = 2 \. Show 
that the locus of their mid-points is 

tf t X 2 [aY + &V] 
a 2 ~*~ 6* + ay + 6V 

33. P and D are the ends of conjugate diameters, and normals at 
these points meet in M. 

Prove OM ± PD. 

34. OP, OD, and OP 7 , OD', are pairs of conjugate semi-diameters. 
Prove A POP' = A DOD'. 

35. P is any point on an ellipse. ±s through P to AT, AP, meet 
A' A in M and N. 

Prove MN = latus rectum. 

36. P and Q are any two points on an ellipse. AP and A'Q meet in 
M, and AT and AQ meet in N. 

Prove MN _L AA'. 

138. The eccentric angles of the extremities of two Conju- 
gate diameters differ by a right angle. 

Let PP' and DD' be two 

conjugate diameters. Let the 
co-ordinates of P be (a cos cf>, 
b sin <£), and the equations of 
OP and OD be y — mx, and 
y = m x x, respectively. 

m1 y b sin <£> 

Then, m = £ = — - , 

x a cos <f> 





/" 


\ 




/ 


K 




/ 


\ 




\ 


/ 


\ ^^^^m^ 




\ 


' £ 


X^X 




p\ 


/ s 


v v 




X \ 


1 s 


>v ^ 




X \ 


1 / 

// 

if 


\ \ 
\ \ 

\ \ 

\\ 




X \ 

\\ 

\l 


\ M X> 


\ M ' l 




^x // 

\ /I 


* x S 


X. .X > 


N X/ 


X. ./ / 


\P*x^ 


^Sh'/ 


\ 






/ 



Fig. 119. 



and 



6 sin <£. 
m = p 

1 » COS <pj 

# 2 sin <£ sin <£ x 



whence 



a' a 2 cos cf> cos ^ 
cos <£ cos (f> t -f- sin <£ sin <£ x = 0. 

•'• <t> L — <£ = 7>' 



.'., etc. 



' THE ELLIPSE 181 

139. OP 2 + OD 2 is constant. — Let the eccentric angle of 
P be <f>, then that of D is <f> + £ . Then P is the point [x lf yj 
or. [a cos <f>, b sin </>], and D is [jc 2 , ?/ 2 ], or 



L cos f <f> + ^j , £ sin U + ^ j 



or, [—a sin <£, 6 cos <£]. 



OP 2 = x 2 + y 2 = a 2 cos 2 <£ + b 2 sin 2 <£, 



and OD 2 = x 2 2 + y 2 2 = a 2 sin 2 <£ + b 2 cos 2 <£. 

.-. OP 2 + OD 2 = a 2 + 6 2 . .••. , etc. 

140. The area of the parallelogram whose sides are tan- 
gents at P, P', D, and D', is constant. — 

For its area = 4 EJ whose adjacent sides are OP, OD, 
= 8 A OPD, 

= 4 (#1^2 - V\ X ^ § l 1 ! COr - 1 ' 

= 4 ab (cos 2 <f> + sin 2 <f>), 
= 4ab. 
.-., etc. 

141. If P is (x x , y^ and D is (x 2 ,y 2 ), to find P' and D'.— 

b 2 x 
The slope of the tangent through P is ~ • 

.\ the diameter DD' parallel to it, is 

b 2 x, 

y = — 2— ^ 

xx, yiL _ 
Now D is on this diameter and also on the ellipse. 

" a 2 ~*~ b 2 W 

and nr + lH 1 (2) 

a 2 b l 

Solving (1) and (2) for x 2 and y 2 , we get, 



182 ANALYTICAL GEOMETRY 



a 
x 2 = ± -y x , 

b 

y, = T-x t . 

The upper signs give the co-ordinates of D, and the lower 
signs those of D' in terms of x t and y v Similarly the co- 
ordinates of P' may be found. 

Note. — Equation (1) expresses the relation that must exist between 
the co-ordinates of the extremities of every pair of conjugate diameters ; 
i. e., if the ends of a diameter be given, the ends of its conjugate diameter 
can be found. 

EXERCISES. 

1. Any tangent to an ellipse meets the tangent at A in K and the 
minor axis in k. Prove, fcK = kF. 

Note. — In the exercises of this chapter, the letters M, N, etc., do not 
mean the points M, N, etc., in Fig. 114, § 118. The letters A, A', F, F', 
however, always stand for the vertices and foci. [0 is center.] 

2. Points L, L', are taken on the minor axis so that 

OL = OL' = OF. Also, p, p', are the ±s from L, L', on any tangent. 
Prove p 2 + p' 2 = 2 a 2 . 

3. ±s from the foci on a pair of conjugate diameters meet in R. Show 
that the locus of R is the ellipse 

b 2 y 2 + a 2 x 2 = a 2 (a 2 -b 2 ). 

4. P is the point of contact of a tangent to the ellipse. The locus of 
the intersection of a _L from center on the tangent at P, with the ordinate 
of P, is the ellipse a 2 x 2 + b 2 y 2 = a 4 . 

5. P, D, are the ends of conjugate diameters, and PD makes an /_ \p 
with major axis, d is ± from O on PD. 

Prove d 2 = % [a 2 cos 2 ^ + b 2 sin 2 rp]. 

6. The line x cos a + y sin a = p will be a normal to the ellipse 

a 2 + b 2 *' n 
p 2 (a 2 sin 2 a + b 2 cos 2 a) = (a 2 — 6 2 ) 2 sin 2 a cos 2 a. 

Note. — Compare given line with equation of normal. 



THE ELLIPSE 183 

7. Prove that if the semi-conjugate diameters OP and OD meet the 
tangent at A in R and S, then 

AR • AS = a constant. 

8. a, /3, are the eccentric angles of the extremities of a chord of a 
series of parallel chords. 

Prove a + j3 = a constant. 

9. P. D, are the ends of two conjugate diameters. A parallel to 
PD meets OP, OD, and the ellipse, in L, M, N, respectively. 

Prove NL 2 + km 2 = PD 2 . 

10. POP', QOQ', are any two diameters, a, /3, the eccentric /j, of P 
and Q. A ^=7 is formed by tangents at P, P', Q, Q'. 

Prove area of & = - — — • 

sin (a — /3) 

11. P is any point on an ellipse. FP = a, and the ± from F on 

tangent at P = a'. 

Prove £_«f + l_0L 

12. The JL from on tangent at P = a, OP = a'. 

a 2 ¥ 



Prove a 2 = 



[a 2 +& 2 -A' 2 ] 



13. POP', DOD', are conjugate diameters of the ellipse 2 + |£ = 1. 
L is any point on the circle a' 2 + y 2 = r 2 . 

Prove LP 2 + LP> 2 + LI? + LD 72 = 4 r 2 + 2 (a 2 + 6 2 ). 

14. MFM' is the latus rectum. Any ordinate PN is prolonged to 
meet the tangent at M in S. 

Prove FP = SN. 

15. FG, FG', are ±s from focus F on a pair of conjugate diameters. 
Prove the line GG' meets the major axis in a fixed point. 

142. Equal conjugate diameters. — If AA' and BB' are 

the axes of an ellipse, then the diameters parallel to AB and 
A'B are equal [by symmetry] and they are conjugate. Hence, 
the equi-con jugate diameters are the diagonals of the rect- 
angle whose sides are tangents to the ellipse at the extremities 
of the axes. 



184 



ANALYTICAL GEOMETRY 



±-x, 
a 



Their equations are y 

To find their length, we have 

OP 2 + OD 2 = a 2 -f& 2 ; 
andli OP = OD = a 19 

theD 2 a 2 = a 2 + b 2 . .-., etc. 



139 



r>s^~~N_ 


— -^^p 


X** 1 




/ "•» ■ 




/ N 1 




/ V 1 




1 XI 


\N' \ 


V ° 


1 / 




V 1 / 








V \ J 








x 1 X 




_^s%' 



Fig. 120. 



143. Supplemental Chords. — The lines joining any point 

of an ellipse to the ends of a 
diameter are called supple- 
mental chords. 

Thus, PQ and PQ' are such 
chords. 

Draw ON II to PQ' and ON' II 
to PQ. 

Then N and N' are the mid- 
points of PQ and PQ'. 
.*. ON will bisect chords parallel to PQ, and ON' will bisect 
chords parallel to PQ'. 

.-. ON and ON' are conjugate diameters. 
This can be easily proved analytically by the student. 
To draw a pair of conjugate diameters which shall include 
a given angle, we proceed thus : 

On AA' describe a segment of a circle to contain the given 
angle, and cutting the ellipse in L and L'. Then the diam- 
eters, parallel to LA and LA' or L'A and L'A' are evidently 
those required. 



144. Poles and polars. — The chord of contact of any 
point (x lf y^) is 



a 2 "*" b 2 



1, 



THE ELLIPSE 185 

The polar of (x lt y^) is 

a 2 "■" b 2 
The polar of the focus (ae, o) is 

a 

x = -, 

6 

.-. tangents at the ends of a focal chord meet in the 
directrix. 

Note. — The above equations are easily found by proceeding in a 
manner analogous to that used in the circle and parabola. The theorems 
and remarks given there hold true here also. 

145. Equation of the ellipse referred to a pair of Con- 
jugate Diameters. — The formulae for transformation are, 

x = x' cos + y' cos 0', 
y = x' sin + y' sin 0', 

where and 0' are the angles made by the diameters [new- 
axes] with the former .r-axis. 

Then, substituting and dropping accents, the equation 

^ + ^ = i 

a- b- 



becomes, 






cos 6 cos 6' sin sin 6'~\ 

+ T2 \ X V 



a- 
sin 2 



cr ' lr 



(1) 



-o . . c , a, sin 6 sin 6 b- 1 Q „ 

But tan • tan = = . § 137. 

cos 6 cos a 2 

sin $ sin 0' cos 6 cos 6' __ 

b 2 a 2 

.-. (1) reduces to, 

Tcos 2 sin 2 #1 . fcos 2 6' sin 2 6"] „ 4 /m 



186 



ANALYTICAL GEOMETRY 



Hence the curve is obliquely symmetrical with respect to 
its new axes. 

If ± a and ± b x are its intercepts on the axes, we have 
1 



= coefficient of x 



coefficient of y 2 , 



. in equation (2). 



Hence the equation may be written 



_ 4_ V_ _ 1 
a 2 ^ b 2 



(3) 



or, when the axes are the equi-con jugate diameters, we have 

x 2 + if = a 2 . 

The equation (3) has the same form as the equation of the 

ellipse referred to its own axes. Hence, any formulas derived 

from the equation 

x * _l_ y 1 _ 1 
a 2 + ¥ ~ > 

which are independent of the angle between the axes, will 
also hold true when the ellipse is referred to a pair of conju- 
gate diameters ; e.£.,the tangent at (x,, y,) referred to the con- 
jugate diameters is 

a* ^ b 2 



Likewise for the chord of contact, polar, etc 
Y 




Fig. 121. 



146. Polar equation 
of the ellipse referred 
to left-hand Focus. 

P is any point (p, 6) on 
r— -, — x the ellipse. 

Now p = a 4- ex, 
and x = ON = FN - FO 
= p cos — ae. 
.♦. p = a + ep cos 1 — ae 2 , 



THE ELLIPSE 187 

whence p = - — ^ — — '— . 

1—6 cos 6 

Discussion : 
When 6 =-- o, p = a + ae = FA'. 

IT 

^ = t>' P = « (1 — e') = semi-latus rectum. 
= 7r, p = a — ae — FA. 
= ^ 7r, p = a (1 — e 2 ) = semi-latus rectum. 
d=2ir, p = a + ae = FA'. 

Summary. — As 6 varies from o to 7r, p decreases from a + ae to a— ae; 
and as 6 varies from -k to 2 7r, p increases from a — ae to a + ae. Since 
e < 1 and cos cannot be > 1, p is always positive. 

Referred to the right-hand focus, the equation of the ellipse is 

a (1 - e 2 ) 

P = — I; • 

r 1 + e cos 6 

The discussion is left to the student. 



EXERCISES. 

1. A focal chord is ± to the line joining its pole to the focus. 

2. FN is any ordinate of an ellipse, and the ± from P on its polar 
meets the major axis in L. 

Prove OL = e 2 • ON. 

3. Show that the pole of the line x cos a + y sin a = p is 

ra 2 cos a o 2 sin a" 



sin a~| 
V J 



L V 
Suggestion. — Compare given line with equation of polar. 

4. Show that the locus of a point whose polars with regard to the 

ellipse 2 + ' = 1 touch the circle x 2 + y 2 •= r 2 , is the ellipse 

x 2 y 2 _ 1 
a 4+ 6 4 ~^' 

5. Tangents to a parabola at P, Q, meet in T. Any other tangent 
meets TP, TQ, in M, N. Show that the locus of the intersection of 
PN and QM is an ellipse touching TP and TQ at P and Q, respectively. 



188 ANALYTICAL GEOMETRY 

6. OP, OD, are conjugate radii of an ellipse, AA', BIT, axes. Prove, 

(1) PA • PA' = DB • DB'. 

(2) The bisectors of /* APA', BDB', are JL. 

7. The product of the three normals that can be drawn from a point 

(A, k) to ellipse — 2 + 9- = 1 is 

2ab(d i -e 2 h 2 )* 



a 2 -b 2 

8. Prove that normals to the ellipse from any point on either of the 
lines a 2 x 2 — b 2 y 2 = meet the curve in points, a pair of whose joins are 
parallel.* 

9. Two supplemental chords of an ellipse, PQ, P'Q, meet the 
tangents at P' and P in T', T, respectively. 

Prove, PT • P' T' is constant. 

10. In Fig. 117, § 135, prove FL' and F'L meet on the mid-point of 
the normal PN. 

11. P and D are the extremities of two conjugate diameters. The 
feet of ±s from center on the tangents at P and D are M, M', and the ±s 
meet the ellipse in N and N'. 

Prove ' =ni — =5 + 2 — 2 = -r + IT • 

OM 2 • ON 2 ^ OM /2 • ON' 2 « 4 bi 

12. Points L,L r , are taken on the normal at P so that PL = PL' = OD. 
Prove OL = a - 6, OL' = a + b. 

Note. — P and D have same meaning as in preceding exercise. 

13. Two tangents to the ellipse from T make an /_ <p. 

FT 2 + IPT 2 — 4 a 2 
Prove, cos = ^ T ^ [where F, F', are foci]. 

EXERCISES ON CHAPTER VIII. 

1. Show that the angle between two conjugate diameters is a maxi- 
mum when they are equal. 

2. Find the polar of the focus with reference to each auxiliary circle. 

Arts, ex = a, ex = — 
a 

Note. — The minor auxiliary circle is x 2 + y 2 = b 2 . 



The join of two points is the line between them. 



THE ELLIPSE 189 

3. Angle between the equi-conjugate diameters of an ellipse is 120°. 
Show, eccentricity = J V6. 

4. The tangent at end of latus rectum is y + ex = a. 

5. Length of chord joining ends of two conjugate diameters is 
Va 2 + b 2 + a 2 e 2 sin 2 </>. Show that its greatest value is a V2. 

Note. — 4> is eccentric /_ of one end ; + - of other end. 

6 Find locus of vertices of ^s constructed on the conjugate diam- 
eters of an ellipse. 

7. Of all sz7s circumscribed about the same ellipse, those constructed 
on two conjugate diameters have a minimum area. 

8. Of all ^s inscribed in same ellipse, those whose diagonals are 
conjugate diameters have a maximum area. 

9. To inscribe the greatest ellipse in a given £J . 

10. To circumscribe the least ellipse about a given C7. 

11. From an external point to draw a tangent to an ellipse. Also a 
normal. 

12. Of all pairs of conjugate diameters of an ellipse, the axes of the 
curve for the least sum, and the equi-conjugate diameters the greatest 
sum. 

13. Any rectangle being circumscribed about an ellipse, show that the 
^7 formed by joining the points of contact of the sides has a constant 
perimeter ; also two consecutive sides make equal angles with the 
tangent. 

14. Find the poles of the directrix of the ellipse with regard to the 



auxiliary circles. 



Ans. (ae, o) 



6*4 



15. In an ellipse, show that the equi-conjugate semi-diameter is to 
the semi-diagonal on the axes as 1 : \/2. 

16. Given an ellipse and x 2 + y 2 = a 2 , find locus of intersection of 
normals drawn to the ellipse and circle at corresponding points. 

Ans. A circle, x 2 + y 2 = (a + b) 2 . 

17. A A is inscribed in an ellipse, d v cl. 2 , e? 3 , are semi-diameters 
parallel to the sides of the A, and R is the radius of its circumscribed 
circle. Prove, 

d x ■ d 2 • d 3 



R = 



ab 



190 ANALYTICAL GEOMETRY 

18. Show that the locus of poles of normal chords of an ellipse is 
x 2 y 2 (a 2 - b 2 ) 2 = a Q ij 2 +b 6 x 2 . 



y 



19. In the ellipse — + f= = 1 

a 2 b 2 

find the diameters conjugate to 



(1) ax-by = 0,\ \a?y+ ¥ x = 0, 



(2) bx — ay = 0.) ' ( ay + bx = 0. 

20. Show that the line x cos a + y sin a = p is a tangent to the ellipse if 

p = Va 2 cos 2 a + b 2 sin 2 a- 

21. A chord PQ parallel to AB meets OA, OB, in p, q. 
Prove, Fp = Qq. 

Also, if PM, QN, are the ordinat.es of P and Q, prove, 
2 AM • AN =~Ap 2 . 

22. Any diameter meets parallels through P, D, to any tangent in p, d 
[where P, D, are ends of conjugate diameters], and the tangent in T. 
Prove, Op 2 + OeP = OT 2 . 

23. Any line through vertex A meets the ellipse in M, and minor axis 
in N. OP is a semi-diameter parallel to AM. 

Prove, AM AN = 2 OP 2 . 

24. P and Q are the points of contact of the tangents to an ellipse 

fromT (h, k). Prove, 

FT 2 
FP • FQ = T ^ T2Y" t F is the focus]. 



2 + b 2 



w 

25. P, Q, R, are any three points on an ellipse. The diameter bisect- 
ing QR meets PQ, PR, and the curve in M, N, and L. 

Prove, OM • ON = OL 2 [O is the center always]. 

26. Find for the locus in Ex. 18, the result 

a 8 b 6 

- 2 + - 2 = a 4 e 4 , 

x l if 

which is the locus of the pole of a normal chord to the ellipse. 

27. The center of a circle of constant radius moves on a diameter of 
the ellipse. Find the locus of the intersection of the common chords of 
the ellipse and the circle. 

28. An ellipse which intersects a fixed straight line revolves about its 
center. Find the locus of the intersection of the tangents to the ellipse 
at its points of meeting with the line. 



THE ELLIPSE 191 

Note. — Turning the axes of co-ordinates through a negative angle is 
the same as turning the ellipse through a positive angle. 

29. Show that the polar of a point on a diameter is parallel to the 
conjugate diameter. 

30. The ratio of the subnormals for corresponding points on the 

2 

ellipse and circle x 2 + y 2 = a 2 is =- • 

31. Pj P 2 is any chord of an ellipse _l_ to A A'. Find locus of the inter- 
section of A'Pj with AP 2 . Ans. An hyperbola, same axes as ellipse. 

32. Show that the J_ from center of an ellipse to a line joining the 
ends of two ± diameters is constant. 

33. A J_ is drawn from focus to a diameter. Find locus of its inter- 
section with the conjugate diameter. 

Ans. Straight line± major-axis. 

34. The semi-major axis of an ellipse is a mean proportional between 
the intercepts on that axis of the lines which join any point of the curve 
to the extremities of the minor axis. 

35. With the co-ordinates of any point of an ellipse as semi-axes, a 
concentric ellipse is described, its axes in same direction as those of 
given ellipse. Prove that the lines joining the extremities of the axes of 
the first ellipse are tangent to the second. 

36. The locus of the mid-point of the chord of contact of two ± tan- 
gents to an ellipse is 

(a 2 + b 2 ) (b 2 x 2 + ahj 2 ) 2 = a*b l (x 2 + y 2 ). 

37. The ± from center of ellipse on tangent at P = — where b' is 
the semi-diameter, conjugate to the diameter through P. 

38. Two conjugate diameters of an ellipse are prolonged to cut the 
tangent at the vertex. Prove that the semi-minor axis is a mean pro- 
portional between the parts of tangent cut off. 

39. The product of the distances of the center of the director circle 
and of any point on it from the polar of the point [with respect to the 
ellipse] is constant. 

40. The ordinate of any point on an ellipse is produced to meet the 
tangent at the end of latus rectum. Show that the length of the ex- 
tended ordinate equals the length of the corresponding focal radius of 
the point. 



192 ANALYTICAL GEOMETRY 

41. Tangents from P to an ellipse make angles d and 0' with major 
axis. Find locus of P if 

(1) Tan d tan 6' = k, a constant. 

Ans. y 2 - W =k (x 2 - a 2 ), an ellipse or hyperbola 
according as k is negative or positive. 

(2) 6 + 6' =± 2 0, a constant. 

Ans. {if - x 2 + a 1 - b 2 ) tan 2 + 2 0:2/ = 0. 

(3) Tan + tan 0' = fc ; a constant. Ans. fcz 2 - 2 a*/ = ka\ 

42. P is any point on auxiliary circle x 2 + y 2 ±= a 2 . The tangent at 
P meets AA' in T. If PA, PA', meet the ellipse in M, N, prove that 
chord MN passes through T. 

43. Also, ±s from foci of ellipse on the tangent at P [Ex. 42] meet 
the ellipse in Q, Q'. 

Trove PQ, PQ', are tangents to the ellipse. 

44. P, Q, are the points of contact of tangents from T (h, k). Prove, 
area of A OPQ [0 is center of ellipse] . 

_ a 2 b 2 Vb 2 h 2 + a 2 k 2 - a 2 b 2 
b 2 h 2 + a 2 k 2 
and area of quadrilateral TPOQ 



= Vb 2 h 2 + a 2 k 2 - a 2 b\ 

45. TP, TQ, are tangents from T so that PQ subtends a right angle 
at the center of ellipse. Show that the locus of T is the ellipse 

t 4. t = I . A 
a* ^6* a 2 ~*~ b 2 ' 

46. The locus of the centers of equilateral As whose sides touch the 

x 2 y 2 
ellipse — + ~ = 1 is the curve 
a 2 b 2 

9 (x 2 + y 2 ) 2 - 2 (5 a 2 + 3 6 2 ) x 2 - 2 (3 a 1 + 5 b 2 ) if + (a 2 - b 2 ) 1 = 0. 

47. Tangents are drawn to the ellipse at two points whose eccentric 
angles differ by f tt or 120°. Show that the locus of their intersection 
is the ellipse 

a 2 * b 2 



THE ELLIPSE 193 

48. A parallelogram is circumscribed about the ellipse — +r? = 1- 

If two opposite vertices trace out the curve 

f(x,y)=0, 
find the curve traced by the other two. 

^ 4-1' £1 = °' 

x 2 i/ 2 
where X 2 = — + |- - 1. 
a 2 b 2 



CHAPTER IX 

THE HYPERBOLA 

147. Definition. — If P is a point on the curve, then 

F'P = e • PM', 

where F' is the focus, PM' the _L from focus to directrix DD'. 
and e > 1. 

148. To find the equation of the hyperbola. 



D 


Y 

M 


M 


D 






B 


I 1 i 


F JA 
/ D 


N 


N' 
B' 


A\ F-' R 
D' >. 



Fig. 122. 

Draw FN _L to directrix. Take points A and A r on FN 

such that 

FA = e • AN (1) 

FA'=e- A'N (2) 

Then A and A' are points on the curve [the vertices]. 
Bisect A A' in 0, and let A A' = 2 a. 

194 



THE HYPERBOLA 195 



Add (1) and (2), we obtain 

2 OF = e • 2 OA. 

.-. OF = ae. 
Subtract (1) from (2), 

2 OA = e ■ 2 ON. 



•. ON=* 



e 

Now, through draw BOB' _L to OA. Take OA and OB 
as axes. 

F is the point (— ae, 0), F' is (ae, 0). 

FT 2 = e 2 • Pi 72 [by definition], 

but FT 2 = (x - ae) 2 + y 2 , 

also, PM' = OR - 0W= x - - • 

e 

.'. (x — aef + y 2 = (gjc — <x) 2 . 

... & ( e 2 _ ^ _ y 2 = a 2 (e 2 - 1) . . . . (1) 

The intercepts of this on the ?/-axis are imaginary and 
equal to ± a V— (e 2 — 1). 

Put a 2 (e 2 - 1) = & 2 , 

and divide (1) by its dexter, 

X 2 7 2 
•'•a 2 -p =1 ( 2 ) 

is the equation of the hyperbola. 

Its intercepts [imaginary] on the ?/-axis may also be written, 



y = ± W-l. 

.*. hyperbola does not meet ?/-axis in real points. 
It is, however, convenient to measure OB = 0B' = b. 



196 ANALYTICAL GEOMETRY 

If in (1) we put x = ae (OF), we get 

' a 2 

2 b 2 

.-. 2 y = = latus rectum 

a 

[double ordinate through the focus]. 

Discussion of (2) : 

(1) Symmetrical to rc-axis [called .-. transverse axis]. 

(2) Symmetrical to ?/-axis [called .•. conjugate axis]. 

(3) Symmetrical to origin [called .-. the center], 

(4) No part of the curve lies between the lines x = ^ a, since for 
values of x between 4- a and — a, y is imaginary. Also, while x increases 
from a to oo, y increases from to oo. Hence the curve has two infinite 
branches lying outside of the lines x — J- a. 

(5) The curve has two foci and two directrices. 

149. The difference between the focal * distances of any 
point on the curve is constant and equal to 2 a. — 

F'P = e • PM' = e • (OR - ON') = ex - a. 
FP = e . PM = e . (OR + ON ) = ex + a. 
... FP - FT = 2 a. 

.-. , etc. 

Note. — The polar equation of the hyperbola with center as pole is 
obtained by putting x = p cos 0, y = p sin 0. 
p 2 cos 2 6 p 2 sin 2 
•'■~tf b 2 = ' 

o 2 b 2 a 2 b 2 



" r b 2 cos 2 - a 2 sin 2 b 2 - (a 2 + b 2 ) sin 2 
[The discussion of this equation is left for the student.] 

150. Construction of the curve. — This is based on the 
principle of § 149. The foci and 2 a are given. 
(1) Mechanical : 



THE HYPERBOLA 



197 



Pivot one end of a ruler about the focus F, and to the 
other end join a string whose length is less than that of the 
ruler by 2 a. Press the string against the ruler by a pencil- 




Fig. 123. 

point P, while turning the ruler about F. The point P will 
describe one branch of the curve, for in every position we 
have PF — PF' = 2 a. For the other branch, interchange the 
fixed end of the ruler and the end of the string. 

(2) By points : 

Bisect F'F in 0. In F'F produced take any point D. Also 
lay off OA = OA' = a. Now, with F' as center and A'D as 




F' A' O A\ F D 



Fig. 124. 




198 ANALYTICAL GEOMETRY 

radius, describe two arcs. Then with F as center, AD as 
radius, describe two arcs cutting the former in P and P'. 
These are evidently points on the curve. By interchanging 
radii, R and B/ are found. After a sufficient number of 
points have thus been found, they are joined by a smooth 
curve. 

Note. — Since the equations of the ellipse and hyperbola differ only in 
the sign of 6 2 , any formula deduced for the ellipse may be changed to 
the corresponding formula for the hyperbola by changing b 2 to — b 2 
or b to 6\/— 1. 

151. The equilateral hyperbola. — This is an hyperbola 
whose transverse and conjugate axes are equal. 

Its equation is, .*. x 2 — y 2 — a 2 . 

Also, since b 2 = a 2 (f - 1) 

[put a for b], .'. e = V2. 

Its relation to the general hyperbola is analogous to the 
relation of the auxiliary circle to the ellipse. 

152. The conjugate hyperbola. — This is the hyperbola 
which has BB' for its transverse axis and A A' for its conju- 
gate axis. To get its equation, put x for y, and a for b f in 
the equation 

a 2 b 2 

2 2 

We obtain —-%-=—!, 

a 2 b 2 

It may also be obtained by turning the axes through an 
angle of 90°, and then putting a for b, and b for a. 

153. The idea of asymptotes. — Let parallels to the axes 
through A, B, B', etc., meet in K and K'. 



THE HYPERBOLA 



199 




Fig. 125. 

P is any point on the curve whose ordinate meets OK in Q. 
Then from the equation of the hyperbola we get, 

y [FN] = - V* 2 - a z .....(!} 



The equation of OK is y = - x. 



PN being always less than QN, we have 

b . 

PQ = - \ x - V^TT^2] 

ab 



(2) 



x -f- Vic 2 — <x 2 

From this it is evident that by taking x large enough, 
PQ may be made less than any assignable quantity. Hence 
the curve continually approaches the lines OK and OK' 

whose equations are y = ±- x\ but never meets them. These 

lines are called the asymptotes of the hyperbora. 

Note. — QN 2 - PN 2 = 6 2 [a constant]. 



200 ANALYTICAL GEOMETRY 

154. Another view of asymptotes. — The line y = mx, 
passing through the center of the hyperbola, meets it in two 
points whose abscissas are, 

ab -\ 



Vfe 2 — a 2 m 2 
— ab 

V6 2 - aV 



(1) 



It also meets the conjugate hyperbola in two points whose 
abscissas are, 

ab 
x t = 



\la 2 m 2 — b 2 

— ab 
*\/a 2 m 2 — b 2 . 



(2) 



Hence the points (1) will be imaginary, situated at infinity, 

or real, according asm 2 >, = , or < — . 

a z 

And the points (2) will be imaginary, situated at infinity, 

b 2 
or real, according as m 2 <, =, or > — . 

a 1 

.-.if a line through the center meet an hyperbola in real 
points, it meets its conjugate in imaginary points and vice 
versa. 

We shall now give a definition of an asymptote ; viz., a 
straight line which passes through finite points and meets a 
curve in two points at infinity. 

.-. if the line meets the hyperbola at infinity, we have 

b 2 b 

m 2 = -= , and the equation of the line becomes y = - x. 
a 2 a 

.-. the hyperbola has two asymptotes [one for each branch], 

whose equations are 



THE HYPERBOLA 201 

b , b x 2 y 2 

y=-x, and y = x or __^L=.0 

a a a? b 2 

Note 1. — If a line through the center meets the conjugate hyperbo- 
las in M and N, it can be shown that OM 2 = - ON 2 or OM = ON V^T. 
Or, if M is (h, k), N is (h V^l, k V^T). 

.-. if M is a real point, N is imaginary, and vice versa. 

Note 2. — Equations of the hyperbola, conjugate hyperbola, and 
asymptotes have identical sinisters ; viz., 



.r 2 
a- 


y 2 

b 2 


= 


+ 


i 


X 2 

a 2 


y 2 
b 2 


= 


- 


1 


X 2 

a 2 


y 2 
b 2 


= 








(1) 

(2) 
(3) 



EXERCISES. 

1. AOB and COD are two straight lines which bisect each other at 
right angles in 0. Find the locus of P which moves so that, 

PA • PB = PC • PD. 
Ans. Rectangular hyperbola, x 2 — y 2 = % (a 2 — 6 ? ), 
where AB = 2 a, CD = 2 b. Take AB, CD, as axes. 

2. The ordinate of a point on a hyperbola is produced until its length 
equals a focal radius of the point. Find the locus of its extremity. 

3. Show that the distances of either focus of an hyperbola from the 
asymptotes equal half the minor axis. 

4. If, from any point of a hyperbola, a line is drawn parallel to 
either axis, the product of the parts cut off between the point and the 
asymptotes equals the square of half that axis. 

5. P is any point on a rectangular hyperbola. 

Prove, FP • F'P = OP 2 . [F, F', are the foci, the center.] 

6. Two straight lines of lengths a and b slide along two rectangular 
axes in such a manner that their extremities are always concyclic. Find 
the locus of the center of the circle. 

Ans. Hyperbola, x 2 — y 2 = \(a l — b 2 ). 



202 ANALYTICAL GEOMETRY 

7. Prove that the foot of a _L from a focus to an asymptote is at dis- 
tances a and b from the center and focus, respectively. 

8. P and Q are any two points on a hyperbola. Parallels through 
P and Q to the asymptotes meet in L and M. Prove that LM passes 
through the center of the hyperbola. 

x 2 y 2 

9. PP' is a chord of the ellipse — -f %- n =A parallel to its minor axis. 

or b l 

Find locus of intersection of AP and AT'. A and A' are vertices. 

x 2 y 2 
Ans. The hyperbola, — , — y — = 1. 
a 2 b 2 

10. The eccentricities of two conjugate hyperbolas are e x and e 2 . 
Prove, 

(i>Vi-i. 

(2) ae x = be 2 . 

11. Prove that the circles described on parallel chords of a rectangular 
hyperbola as diameters are coaxial. 

12. A line through the center of an hyperbola meets the curve in P, 
and the lines drawn through the vertex A parallel to the asymptotes, in 
T, T'. Prove, 

OP 2 = OT • OT/. 
13. The asymptotes meet the directrices of an hyperbola on the 
auxiliary circle x 2 +y 2 =a 2 , and those of its conjugate hyperbola on the 
circle x 2 + y 2 = b 2 . 

x 2 y 2 
14. The equi-conjugate diameters of the ellipse — - -f- ^ = 1 coincide 

x 2 y 2 
with the asymptotes of the hyperbola — — ^ = 1. 



V 



15. A circle with center at a focus and radius = b, passes through the 
intersections of the asymptotes and the corresponding directrix. 

155. Miscellaneous facts. — Most of the work of the pre- 
ceding chapter is applicable to the hyperbola if b is changed 
to b V— 1, thus : 

(1) The tangent at (x x , y t ) is 

x ll - y ll = 1 
a 2 ¥ 



THE HYPERBOLA 203 

(2) The normal is 



b 2 X, 



(3) The subtangent is == — • 

x i 
b 2 x 

(4) The subnormal = -~ • 

(5) The tangent and normal at any point bisect the angles 
between its focal radii. 

(6) The line y = mx ± \Jm 2 a 2 — b 2 is a tangent to the 
hyperbola for all values of m. 

(7) The director circle is x 2 + y 2 = a 2 — b 2 . 

(8) The chord of contact of (x l} y t ) is 

a^i _ yjh = -, 

a 2 b 2 



(9) The polar of (x lf y x ) is 

7 = 1. 



,2 7,2 



a" b 2 

(10) Two distinct, coincident or no tangents to an hyperbola 
can be drawn from a given point according as it is without, 
on, or within the curve. 

Note 1. — The geometrical properties of the ellipse can be proved for 
the hyperbola in an analogous way. 

Note 2. — The director circle of the hyperbola is a point circle if 
a = 6, and is imaginary if b > a. 

156. Equation of diameter. Conjugate diameters. — The 

equation of the diameter is found in the same way as in the 

b 2 
ellipse, and is written ?/ = - 7 — x, where m is the slope of the 

a 2 ?n 

chords. 

If y = m x x is another diameter which bisects all chords 

b 2 b 2 

parallel to the first, we have m. = — — or ram, = — : and if 

1 a 2 m l a 2 



204 



ANALYTICAL GEOMETRY 



the first bisects all chords parallel to the second, we get the 

same relation. 

.*. If one diameter bisects all chords parallel to another, the 

second bisects the chords parallel to the first. Two such 

diameters are called conjugate diameters ; i.e., they go in pairs. 

b 2 
Discussion. — (1) The relation mm, = — shows that the slopes of 

two conjugate diameters are like in sign, hence two such diameters lie 
on the same side of the conjugate axis of the hyperbola, and their 
included angle is .-. acute. 

(2) If m < - i m 1 > - ; but the semi-angle between the asymptotes 
= tan -1 - • .-.of two conjugate diameters, only one meets the curve 
in real points. 

(3) The relation may also be written ( — ) ( — J = rj • 

But this is the condition that the lines y = mx and y = m^x, may be 

x 2 %ft 
conjugate with respect to — — ^ = — 1. 
a- o' 

Hence, two lines which are conjugate diameters of an hyperbola are 
also conjugate diameters of its conjugate hyperbola. 

157. Eccentric angle. Other properties of the hyperbola. 

— Tangents to two conjugate hyperbolas at the extremities of 



\ ^N?— ~~ 




•-"' / /if 
/ / '/ 


\ / \ 

\ / 
1 / 


\0 




4/ _^ 




\ ,'U 


/ / / 




r ~- — ~~M\ \ 


/i / /,i,— 






Ak^^ 







Fig. 126. 



THE HYPERBOLA 205 

two conjugate diameters form a parallelogram whose diag- 
onals coincide with the asymptotes. 

Let PP' and DD' be conjugate diameters. Then it is suffi- 
cient to show that the tangents at P and D form with OP and 
OD a parallelogram whose diagonal OL is on one asymptote. 
Let the co-ordinates of P be 

x = a sec <f>) 

y = b tan <f>) 
where <f> is the eccentric angle. These values evidently sat- 
isfy the equation of the hyperbola. Then the co-ordinates of 
I) are 

x = a tan <f>) 

y = b sec <£) ' 

For, (a) these values satisfy the equation of the conjugate 

hyperbola, 

and (/?) the lines OP and OD, whose equations are 

b tan <£ 
?/ = x (3) 

J a sec <f> v J 

and ?/ = x (4) 

J a tan cf> w 

satisfy the condition for conjugate diameters, viz., 

b 2 
mm, — — • 
1 a 2 

Now, the equation of the tangent at P is 

x II 

- sec <f> — j tan <f> = 1 (5) 

Note. — This is easily found by substituting the co-ordinates of P in 

a- b 2 
[This line is parallel to (4).] 
Also, the tangent at D is 

x 1/ 

— tan <b — ^-sec d> = — 1 (6) 

a b 



206 ANALYTICAL GEOMETRY 

[This line is parallel to (3).] 
Adding (5) and (6) we get, 



(1 ~ I ) (S6C * + tan & = °' 



35 




[ y 


CI 




b ' 
b 


• 2/ 


— 


-X. 

a 



or, 



,-. The lines (5) and (6) meet on the asymptotes. .-., etc. 

158. The difference between the squares of two conju- 
gate semi-diameters is constant. 

From equations (1) and (2), we have, 

OP 2 = a 2 sec 2 <f> + b 2 tan 2 <f>, 
OD 2 = a 2 tan 2 cf> + b 2 sec 2 <f>. 
.-. OP 2 - OD 2 = (a 2 - b 2 ) (sec 2 <£ - tan 2 <f>) 
= a 2 — b 2 . .-., etc. 

159. The area of the parallelogram MLES is constant. 
O MLRS = 8 A POD 

= 4 (a sec cf> • b sec <£ —a tan <£ • b tan <f>) 
= 4 ab. .-., etc. 

Note. — It can be easily shown, both geometrically and analytically, 
that PD is bisected by one asymptote and is parallel to the other. 

160. The portion of a tangent intercepted by the asymp- 
totes is bisected at the point of contact. 

For PL = OD = OD' = PM. 

.-. LM is bisected at P. 

Note 1. — If is the angle between OP and OD, and OP = a,, OD 

= 6, , then sin = — — . 
a A 
Note 2. — Tangents at the extremities of any chord meet on the 

diameter of that chord. 

Note 3. — The polar of a focus is the corresponding directrix. 



THE HYPERBOLA 207 



EXERCISES. 



1„ A chord of a rectangular hyperbola subtends angles at the ex- 
tremities of any diameter which are either equal or supplementary. 

2. TP and TQ are tangents from T. A parallel to an asymptote 
drawn through T meets PQ in R. Show that TR is bisected by the 
[hyperbola] curve. 

3. P is any point on a rectangular hyperbola. OL is ± to the tan- 
gent at P. Prove, 

OP • OL = a 2 . 

4. A ± from center on any tangent to the hyperbola x 2 — y 2 = a 2 
meets the tangent in L and the curve in L'. 

Prove, OL • OL' = a\ 

5. An ellipse and an hyperbola have the same axes. Show that the 
polar of any point on either curve is a tangent to the other. 

6. In two conjugate hyperbolas, show that the polar of the vertex 
of one hyperbola with respect to its conjugate, is the tangent at the 
other vertex. 

7. Conjugate diameters of an equilateral hyperbola are equal. 

8. The parts of a normal to an hyperbola cut off by the transverse 
and conjugate axes, respectively, are in the ratio of b 2 : a 2 . 

9. Every point [except the center of a coipc] has a definite polar 
with respect to the conic, and conversely, every line [excepting one 
through the center] has a definite pole. 

10. The polars of a point with regard to two conjugate hyperbolas 
are parallel. 

11. In an equilateral hyperbola, the length of a normal at any point 
equals the distance of that point from the center. 

12. The line x cos a + y sin a = p, is normal to the hyperbola 

x 2 _y 2 _* 
a* b 2 ~ ' 
if p 2 (a 2 sin 2 a — b 2 cos 2 a) = (a 2 + b 2 ) 2 • sin 2 a cos 2 a. 

13. The tangents to an equilateral hyperbola at the positive ends of 
the latus rectum are 

y = _|_ x v x 2 - a. 

14. Show that the segments of any line which are intercepted between 
an hyperbola and its asymptotes are equal. 



208 ANALYTICAL GEOMETRY 

15. Show that tangents from the foot of the directrix to an hyper- 
bola meet the curve at the extremities of the latus rectum. And, if <p 
is the angle between them, 

Prove, tan = _J_ e . 

16. Tangents to an hyperbola are drawn from any point on the con- 
jugate curve. Show that their chord of contact will touch the opposite 
branch of the conjugate curve. 

17. is the angle between the asymptotes of an hyperbola, e the 
eccentricity of the curve. 

Prove, tan0 = 2 V 



2 — e 2 

18. P is any point on a rectangular hyperbola. Prove, angles PAA' 
and PA' A differ by - ; and the bisectors of /_ A PA' are parallel to the 
asymptotes. 

19. PQ is a chord of a rectangular hyperbola normal at P. Show 
that PQ varies as OP 3 . [PQ oo OP 3 .] 

20. From points on the circle x 2 + y 7 = a 2 tangents are drawn to the 
hyperbola x 2 — y 2 = a 2 . Prove that the locus of the mid-point of the 
chord of contact is the curve 

(x 2 - y 2 ) 2 = a 2 (x 2 + y 2 ). 

21. In an equilateral hyperbola show that focal chords parallel to 
conjugate diameters are equal. 

22. Show that the angles between any pair of conjugate diameters of 
an equilateral hyperbola are bisected by the asymptotes. 

23. P, Q, P, are any three points on an equilateral hyperbola. If 
/_ PQR = 90°, prove that the normal at Q is parallel to PR. 

24. A circle, with a focus [of any hyperbola] as a center and radius 
= b, will touch the asymptotes at the points where they meet the cor- 
responding directrix. 

25. Any tangent to an hyperbola meets the asymptotes in C, C. 
Prove, C, C, F, F' [foci], are concylic. 

26. The segment of a tangent to an hyperbola intercepted by the con- 
jugate hyperbola, is bisected at the point of contact. 



THE HYPERBOLA 209 

161. The asymptote is a tangent from the center. — If 

the tangent y = mx ± y/m 2 a 2 — b 2 pass through the center, its 
intercept is zero. 

.-. s/m 2 a? -b 2 = 0. 



whence 



b 

m = ± - 

a 



Hence the equation of the tangent becomes 

y = ± -x; i.e., an asymptote. .•. , etc. 



162. Equation of the hyperbola referred to its asymp- 
totes. 

r 




Fig. 127. 

OX 7 and OY' are the asymptotes, <£ is the semi-angle be- 
tween them. P is (x, y) to the former axes, (x r , y r ) to the 
new axes. 

x = ON -f NL + LQ = OR cos <f> + RL cos <£ + LP cos <j> 
= (OR + PR) cos <£, 
.*. x = (V -f- y f ) cos <j>. 
y = PQ = PL sin <j> = (PR - LR) sin <£ = (PR - OR) sin <f>. 

•*• y = {y r — x ') sin <£• 



210 



ANALYTICAL GEOMETRY 



.-. Substituting these values in the equation 



x 2 y 2 



we obtain 




— — — 1 



(x' + j/Y cos 2 cf> (y r - x') 2 sin 2 <f> _ 
a 2 ¥ ~ 



Also, since tan <& = — , 
a 



sin <£ = 



cos (f> 



-J a 2 -h 6 2 
a 



Va 2 + 6 2 

Substituting also these values above, we finally obtain, after 
dropping accents, 

4 xy = a 2 -|- b 2 

as the required equation. 
The conjugate hyperbola is 

4 xy = - (a 2 + b 2 ). 

Note. — In the figure of § 159, if the co-ordinates of P, referred to 
the asymptotes, are (x, y), those of D are (— x, y). 



163. The product of the _Ls from any point of an hyper- 
bola on the asymptotes is constant. — Let P [Fig. 127, 

§ 162] be the given point, d l its _L distance from OX', d 2 the _L 
on OY\ 

Then, d x = PR sin 2 <j> = y' sin 2 <£. 

d % = x r sin 2 cf>, 



or, 



d t - d 2 = xy • sin 2 <f> [after dropping accents] 
= a constant 



[since 



xy = a constant]. .*., etc. 



THE HYPERBOLA 211 

164. Equation of a hyperbola with given asymptotes 

A lX + B& + C, = 0, A 2 x + B 2 y + C 2 = 0. 

By the preceding §, if (x, y) is any point on the required 
hyperbola, we have 

K x x + B x y + C x A 2 x + B 2 ;/ + C 2 
1 , - — — - • , = some constant, 

VA X 2 + B* VA 2 2 + B 2 2 

or, (A t a: + Bj?/ -f C t ) (A^a; -f B 2 ?/ -f C 2 ) = a constant A: 
is the equation of the hyperbola. 

Note. — It differs from the equation of the asymptotes which is, 
(A x x -f B t ?/ + C t ) (A 2 x + B 2 ?/ + C 2 ) = 0, by some constant. 

And the equation of the conjugate hyperbola is (A x x + B x y + C x ) 
(A 2 x + B 2 y + C 2 ) = - k. 

Hence, whatever the axes of co-ordinates may be, the two conjugate 
hyperbolas differ only in their constant [numerical] terms, whose values 
are equal and opposite in sign. 

165. The equation of the tangent at (x x , y x ) to the hyper- 
bola xy = k 2 . — Let (x 2 , y 2 ) be an adjacent point of the curve. 
Then the chord is 

y - Vx _ Vi - 2/2 



•V U/i *l/t <b% 



(1) 



but x x y t = k 2 (2) 

z 2 ?/2 = h (3) 

/C AC , 2 

.'. Vx ~ V<1 = [m] = x x ~ x 2 = — 

.*. (1) becomes — — — = ■ [equation of secant]. 

Putting x 2 = x x , 

qj 11 J»2 

we obtain, - ^ = — — - [equation of tangent], 

•AS Js-t JOi 

which may be written in either of the two forms, 



212 ANALYTICAL GEOMETRY 

or ^2/i+ XiV = %k 2 . 

Note. — The equation of the hyperbola whose asymptotes are 

x — 2 y + 1 = 0, 

z+3;/ + 2 = 0, 

is (x-2y + l) (x+3y + 2) + k = 0, 

and conjugate hyperbola is (...)(...)— & = 0. 

Hence another condition is required to determine k. 
Also, the asymptotes of the hyperbola 

x 2 + 2xy -y 2 + Sx + 4y -8 = 

are x 2 + 2 xy - y 2 + 8 x + 4 y + & = .... (1) 

where /c is determined by the condition that (1) may represent two 
straight lines [see § 48] . 

EXERCISES. 

1. A circle meets the hyperbola xy = k 2 in four points, (as,, 2/,), (x 2 , y 2 ) y 

(x-s,y 3 ), (s 4 > S/J- 
Prove, ft x 2 x 3 x 4 = k* = y x y 2 y 3 y 4 . 

Suggestion. — Express that points are concylic. Also each lies on 
hyperbola, etc. 

2. Show that any tangent to the hyperbola x 2 — y 2 = 2 a 2 is cut har. 
monicaliy by the hyperbolas 



y 2 + xy = - c 

Suggestion. —Find equation of tangent to given hyperbola in terms 
of slope. 

Note. — On harmonic division of a straight line. The line AC is said 
to be divided harmonically in B, D, if it is divided internally at B and 
externally at D in the same ratio. Thus, 



B 

Fig. 129. 

AB AD m 
BC ~ DC ~ n 



THE HYPERBOLA 213 

The name harmonic comes from the fact that AB, AC, AD, are in 

harmonic progression. Thus, 

put AB = a, AC = b, AD = c, 

AB AD a c 

then BC = CD' ■■• b^a^^b' 

.-. a (c — b) = c (b — a). 

^. -, , 1111 111 

Divide by abc, .-. T = r- •'• - » r > - » are in A. P. . \,etc. 

b c a b abc 

It may also be shown that DC, DB, DA, are in H. P. Again, if M is 

the mid-point of AC, then we may readily show that 



r 2 u 2 
3. A series of chords of the hyperbola — — ~ = 1 are tangents to the 

circle described on the distance between the foci as diameter. Show 
that the locus of their poles with respect to the hyperbola is the ellipse 
x 2 y 2 1 



a 4 6 4 a 2 + b 2 

4. The pole of any tangent to the rectangular hyperbola xy = k 2 with 
respect to the circle x 2 -\- y 2 = a 2 , lies on a concentric rectangular hyper- 
bola with axes coincident with those of former. 

5. The polar of any point T is parallel to the lines joining the points 
of intersection of the tangents from T with the asymptotes. 

6. Show that the tangents to the hyperbola xy = k 2 at the points 
(x v y t ) and (x 2 , y 2 ) meet at the point, 

r 2 - r i x 2 1 2 vi V2 1 . 

Lx, + x 2 ' y, + y 2 _\ 

7. Show that the asymptotes of the hyperbola 

xy = hx -\- ky are x — k = 0, y — h = 0, 
and its conjugate hyperbola is 

xy = hx + ky — 2 hk. 

8. The asymptotes of the hyperbola 2 xy =hy + kx are 

2.r =/?.) 
2 2/ = k. ] 

9. PM, PN, are _Ls from P on two fixed lines meeting at O, and area 
of quadrilateral OMPN = k 2 [a constant]. If the angle between the lines 
is 0, show that the locus of P, the given lines as axes, is the hyperbola 

4 k 2 



x 2 + 2 xy sec <f> + y 2 = 



sin 2 



214 ANALYTICAL GEOMETRY 

10. A line cuts off a A of constant area = k 2 from two fixed lines, and 
has its extremities in the given fixed lines [of the preceding exercise]. 
Show that locus of its mid-point, taking the given lines as axes, is the 
hyperbola 

4 xy = k 2 . 

Also show that the locus of the centroid of the A is the hyperbola 

_ fr 2 
Xy ~ 9 sin ' 

11. Show that the asymptotes of the hyperbola in Ex. 9 are 

v 

- = — sec + tan 0, 
x 

v 

- = — sec — tan 0. 
x 

12. Show that the locus of poles of normal chords of the hyperbola 
x 2 — y 2 = a 2 , is the curve, 

a 2 (y 2 — x 2 ) = 4 x 2 y ! . 

13. A line has its extremities in two fixed rectangular axes and passes 

through a fixed point. Show that the locus of its mid-point is an 

hyperbola. 

Ans. 2 XV) = hy + kx, where (h, k) is fixed point given. 

14. Find the equation of a tangent to the hyperbola xy = hx + ky 
at (x v y x ) . 

15. The polar of one end of a diameter of an hyperbola, with regard 
to its conjugate hyperbola, is the tangent at the other end of the given 
diameter. 

16. The distance of any point from the center of a rectangular hyper- 
bola varies inversely as the distance of its polar from the center. 

17. In an equilateral hyperbola, lines from any point of the curve to 
the extremities of a diameter make equal angles with the asymptotes. 

18. Show that the line y = mx + — will touch the hyperbola 

x% _ t. _ i 
a 2 b 2 ~ l ' 

if p 2 = m 2 (a 2 m 7 — b 2 ). 

Hence, find the equation of the common tangent to the two curves. 

19. The parts of any chord of an hyperbola intercepted between tht 
curve and its conjugate are equal. 



THE HYPERBOLA 215 

20. Show that the foci of the hyperbola xy = k 2 are the points 
[± fc sec - , ±k sec -J 
where <f> is /_ between axes [asymptotes]. 

166. The portions of any secant of an hyperbola inter- 
cepted between the curve and the asymptotes are equal. 




Fig. 130. 

To prove AA' = BB'. Let A be (x x , y t ), B (x 2 , ?/ 2 ). Then 
the chord AB meets the [axes] asymptotes in B' (x l + ^2?°) 
and A' (p, y x + y 2 ). 

Hence, the mid-point of AB coincides with that of A' B'. 
.-. , etc. 

167. The intercept of any tangent, between the asymp- 
totes, is bisected at the point of contact. — Let P (x l9 y^) be 

the point of contact, Then the equation of the tangent TT' is 

^+^ = 2. 

Its intercepts on the axes are 2 x t , 2y lm Hence the mid- 
point is (x lf 2/1) ; i-e., the point of contact. 
.*., etc. 



216 ANALYTICAL GEOMETRY 

168. The area of the triangle formed by any tangent and 
asymptotes is constant. — 

A TOT' = i OT • OT' sin 2 </>, where 2 <f> is the angle between 
the asymptotes [axes]. 

= ^x±y x sin <f> cos <f> [§ 167] 

= (a 2 + o 2 ) sin <£ cos d> 

7 Va 2 + 7> 2 Va 2 + o 2 
= a6 (a constant). 
.-., etc. 

169. Equations referred to conjugate diameters. — These 

are found in a manner analogous to that in the ellipse. 

3/ 2 "Zy 2 

— a — ~ = 1 [equation of hyperbola]. 

3J 2 I/ 2 

— i — r-g- = — 1 [conjugate hyperbola]. 
f« - fl = ° [asymptotes]. 

170. Polar Equation, referred to left-hand focus. — 

P = ex + a (1) 

# = p cos 6 — c = p cos — ae (2) 

.'. P = V n — L from (1) and (2). 
e cos — 1 v/ w 

The discussion is left to the student. 

Note. — p is positive or negative according as the point («, 6) is on 
the right or left branch of the hyperbola. 

EXERCISES. 

1. Q is any point on a rectangular hyperbola, POP' any diameter. 
Show that the bisectors of ^/PQP' are parallel to the asymptotes. 

2. A tangent to an hyperbola at P meets the conjugate curve in Q 
and Q'. M is the mid-point of PQ, and OM meets the hyperbola in S. 
Prove, OQ, OS, are conjugate diameters. 



THE HYPERBOLA 217 

3. In a rectangular hyperbola two _L diameters are equal. 
Suggestion. — Show they are conjugate. 

4. If two tangents are drawn from an external point, they will touch 
the same or opposite branches of the curve according as the point lies 
between or outside the asymptotes. 

5. The latus rectum of an hyperbola is a third proportional to the 
two axes. 

x 2 v 2 

6. A rectangular hyperbola cuts the ellipse — + ^ = 1 at an angle \p, 



lr 



and its asymptotes are the axes of the ellipse. Find its equation. 

a 2 b 2 cos \p 



Ans. xy = 



v [a 2 + b 2 ] 2 sin 2 \p + 4 a 2 b 2 cos 2 \j/ 
Suggestion. — Assume xy = k 2 for required hyperbola. Find angle 
of intersection with ellipse. Put it equal to \p. Determine k. 

7. Q, R, are fixed points on an hyperbola ; and P, a variable point 
on the carve. PQ, PR, meet an asymptote in G, G'. Show that GG' is 
constant. 

8. 0j and <j> 2 are the eccentric angles of the ends of two conjugate 
diameters of an hyperbola. 

Prove, t + <p 2 = ^ • 

[See Appendix]. 

9. A semi-diameter of an ellipse or an hyperbola is a mean propor- 
tional between the lines drawn from the foci to the end of its conjugate 
diameter. 

10. The director © of the hyperbola xy = k 2 is 

rc ? + 2 xy cos + y 2 = 4 & 2 cos <p [</> is ^/bet • axes]. 
Suggestion. — Show that the equation of the two tangents from the 
external point (x„ y^) is 

[_x x y - y x x] 2 + 4 A; 2 (x - .r,) (y - y x ) = 0. 
Express condition that these lines be _L, etc. 

EXERCISES ON CHAPTER IX. 

1 . The line y = mx + 2k V-wis a tangent to the hyperbola xy = k 2 
for all values of m. The point of contact is 



218 ANALYTICAL GEOMETRY 

2. To draw a tangent to an hyperbola from a given external point. 
Construct an hyperbola, given : 

3. Three of its points and the directions of its asymptotes. 

4. One point, a vertex, and an asymptote. 

5. Find locus of foci of a rectangular hyperbola which has a fixed 
center and passes through a fixed point. Also, locus of its vertices. 

6. Find the locus of center of an equilateral hyperbola required to 
pass through two given points. 

7. Find locus of centers of circles which intercept given lengths on 
the sides of a given angle. Ans. An hyperbola. 

8. Given base of a A and difference of base ^/s = ^ , find locus of 
vertex. Ans. An hyperbola. 

9. Given two points, A, B, find locus of P such that the bisector of 
^/APB may have a given direction. 

10. Every chord of an hyperbola bisects the portion of one or the 
othei asymptote included between the tangents at its extremities. 

11. Find locus of foci of hyperbola, given one asymptote and one 
directrix. 

12. Show that the sum of two focal chords of an hyperbola, which 

2 (a 2 + b 2 ) 

are parallel to two conjugate diameters, is — - • 

a 

13. A circle meets a given rectangular hyperbola in four points. If 
one of the common chords is a diameter of the circle, prove that another 
is a diameter of the hyperbola. 

14. ABC is a given right A, hypothenuse AB. A series of parabolas 
have AB for a common chord, while their axes are parallel to AC. Prove 
that all their foci lie on an hyperbola whose foci are A and B. And the 
hyperbola will be equilateral if A ABC is isosceles. 

15. An hyperbola has a given focus, and intersects in a given point a 
line parallel to one of its asymptotes. Find the locus of its center. 

16. Show that every conic passing through the intersections of two 
rectangular hyperbolas is a rectangular hyperbola. 

17. Prove that every rectangular hyperbola which passes through the 
vertices of a A passes through its orthocenter. 

18. Find the locus of the centers of rectangular hyperbolas passing 
through the vertices of a A. Ans. The nine-points circle of the A. 



CHAPTER X 
CONFOCAL CONICS 

171. Equation of a system of confocals. — The co-ordinates 
of the foci of the conic 

S + p =i w 

are (0. ± Va 2 — &-). These remain unchanged if we put 
a 2 — X, b 2 — X, for a 2 and 6 2 , respectively. Hence the equation 

A + A = 1 v 

will represent any one of a system of confocals if different 
values be assigned to the variable parameter A.. Also if 
\ > b 2 (2) will represent an hyperbola. 

172. Snaps of a confocal. — 

Let a 2 - b 2 = c 2 , a 2 - X = a x 2 , b 2 - X = b 2 . 

.-. a x and b± are the semi-axes of the confocal (2), and we 

have also n 2 , „ 2 , „ „ 

a{ — b{ = a 2 — b 2 = c 2 , 

Then (2) may be written, 

- + £ = i 

6 7T^ + V = 1 (3) 

Discussion of (3) : 

(a) if 6,- is very small and positive, a, 2 = c 2 nearly, .-. the confocal is 
a thin ellipse nearly coincident with the line joining its foci. 

219 



220 ANALYTICAL GEOMETRY 

(/3) If bj 2 is very small and negative, the confocal is a thin hyperbola 
nearly coincident with the a>axis. 

(7) If b 2 — 0, or a = 6 2 , the confocal is either a line-ellipse or a line- 
hyperbola. 

173. Two conies of a confocal system can be drawn 
through a given point. — Let the confocal pass through the 
point (x 19 2/0, Then, 

b*+<*^ bs~ ' 

whence, bf — (x* + y 2 — c 2 ) b* — y 2 c 2 = 0. 

This is a quadratic in b 2 , of which the roots are both real, 
one positive and one negative. 
.-. , etc. 

Note. — One of the confocals is an ellipse, the other an hyperbola. 

174, Elliptic co-ordinates. — The confocal [§ 172] may 
also be written, 

a{ a* — & 
and if this pass through the point (i p y t ), we get 

< ~ 0*V + V x + c ") ci{ + x{C 2 = . . . (4) 
If a 2 , a 2 , be the roots of this equation, we have 

a* • a 2 = X{dr (5) 

[By § 173] similarly, b* • b. 2 = - y 2 c 2 (6) 

= (a 1 2 -c 2 )(a 2 2 - C 2 ). 

The semi-major-axes of the two confocals passing through 
a given point are called its elliptic co-ordinates.* 

Note. — Equations (5) and (6) give the rectangular co-ordinates (ar„ ?/,) 
in terms of its elliptic co-ordinates a x and a,. Also from (4), a x 2 + a 2 2 
= x 2 4- y 2 + c 2 . 



* By Lanie\ 



1. Find the equation of the director circle of the ellipse — 2 + ^ = 1 



CONFOCAL CONICS 221 

EXERCISES. 

p nf t.hp. plli-nsft 

if the elliptic co-ordinates of any point on it are a v a 2 . 

Ans. a 2 + a 2 = 2a 2 . 

2. Tangents are drawn to the same ellipse from a point whose elliptic 
co-ordinates are (a v a 2 ). Show that the angle \f/ between them is found 

from the equation, tan £ \p = a a ~ q z, 

y a 2 — a 1 

or, 4> = 2 tan -1 * / q2 ~ a 2 2 

3. Two parabolas have a common focus. Find the locus of the inter- 
section of two tangents [one to each], which cut at a constant angle. 

Ans. A parabola. 

4. Find the locus of the extremities of the latera-recta of two par- 
abolas which have the same focus aDd a common tangent. 

Ans. Two circles. 

5. Two ellipses have a common focus. One revolves about this focus, 
While the other remains fixed. Find the locus of the intersection of their 
common tangents. Ans. A circle. 

175. Confocals intersect orthogonally . — Let the conies 

i + y i = x o 

meet in the point (x lf y^. Then the tangents to these are 

A + S^T 1 (*> 

Also, since (x t , y t ) lies on each conic, we have 

a 2 + 6 2 " lj 
a 2 - A ^ 6 2 - A 



222 ANALYTICAL GEOMETRY 

By subtraction, 



x * a 2/l = o 

I 79/19 v \ v « 



a 2 (a 2 -A) ' b 2 (b 2 -\) 

But this is evidently the condition that the tangents (3) 
and (4) intersect at right angles. 

.*. etc. 

Note 1. The student can easily prove this geometrically. 

Note 2. If two confocals [whose foci are F and F'] meet in P [ortho- 
gonally], then one of the curves is an ellipse, major axis = FP + FT; 
the other an hyperbola, transverse axis = FP — FT. 

Note 3. The elliptic co-ordinates of P are a x = \ [FP + FT], and 
a 2 - * [FP - FT]. 

EXERCISES. 

1. Show that the locus of the intersection of rectangular tangents to 
the confocal parabolas y 2 = 4 a (x + a), and y 2 = 4 b (x + 6), is the line 
x + a + b = 0. 

2. Show that the above parabolas cut orthogonally [in two points] at 
a finite distance, and that these points are imaginary if a and b have the 
same sign. 

3. Tangents are drawn to a confocal system from a point in the major 
axis. Find the locus of the points of contact. Ans. A circle. 

4. Given three confocal ellipses. From any point P of the outermost 

curve, tangents are drawn to the other two, making angles <p and X with 

the tangent at P. 

■ sin 

Prove, = constant. 

' sm 0j 

N OTE . — The other two confocals may also be hyperbolas. 

5. If a. and \ are the parameters of the confocals which pass through 

x 2 y 2 
any point on the directrix of the ellipse — + ^ = 1, 

prove, aa, = a 2 (a + aJ. 

6. Two conies have one focus in common. Prove that two of their 
common chords pass through the intersection of their directrices. 



CONFOCAL CONICS 223 

176. To find the locus of a pole of a given line with 
respect to a system of confocals. — Let the given line be 

i+ih 1 ci) 

Now, if (x i9 y t ) is the pole of (1) with respect to 

^A^ 1 < 2 > 

itspolaris _^_ + J^.= l (3 ) 

Now, the condition that (1) and (3) may represent the same 
line is A 

-.=A » 

aad I-P?a (5) 

From (4) and (5), we find for the required locus [after drop- 
ping accents], the line, 

hx — ky = a 2 — b 2 , 

which is evidently _L to the given line. 

177. There is one conic and only one of a confocal system 
which can be drawn tangent to a given line. — Let the given 
line be y = mx + c. Its points of intersection with 

_iL +_£_ =i 

o? - \ b- - X 

is determined by -= + -^ ~- = 1 (1) 

ar — A tr — A 

We may now write the condition that (1) may have equal 
roots ; i.e., that the given line may be a tangent. This gives 
an equation of the first degree in A , hence A can only have one 
value. .*., etc. 



224 ANALYTICAL GEOMETRY 

Or, if the given line be 

x cos a -f- y sin a = p, 
the condition that it touch the conic 

* 2 +^ = i 



a 2 - A b 2 - A 
is (a 2 - A) cos 2 a + (b 2 - A) sin 2 a = f. [§ 134.] 

This gives also, an expression of the first degree in A. 



Note 1. — In § 176, there is one conic of the confocal system which 
touches the given line. If this conic be determined, the point of contact 
is found to lie on the locus found in § 176. Hence the required locus of 
§ 176 is the normal to this conic at the point of contact. 

Note 2. — If the tangent at a point P of a conic, cut a confocal in A 
and B, the tangents at A and B intersect on the normal at P. 

EXERCISES. 

1. Tangents are drawn to an ellipse from a point whose elliptic co- 
ordinates are a l and a 2 . Also, normals are drawn to the two confocals 
through the given point [from that point]. Find the equation of the 
tangents referred to these normals as axes. 

X 2 y 2 

(a r 2 — a 2 ) (a 2 — a 2 ) 

2. A circle is described with the focus F of a conic as center. A line 
through F meets the circle in P and the conic in Q. Show that the tan- 
gents at P and Q meet on a common chord of the conic and circle. 

178. To find the locus of the intersection of tangents to 
two confocals, if these tangents meet at right angles. 

The tangent to _ + ^- = 1 

a 2 b 2 

is y — mx = \lm 2 a 2 + b 2 (1) 

The tangent to the confocal 

^ , y 2 -, 

a 2 - A "•" b 2 - A 



CONFOCAL CONICS 225 



is y — m x x = \im 2 (a 2 — A) -+- (b 2 — A) ; 

and if this be _L to (1), m l = , and its equation becomes 



in 



my + x = V(> 2 — A) + i» , (4»- A) ... (2) 

Squaring and adding (1) and (2), we obtain the required 

locus; viz., a' + J^rf+P-A. 

Note. — This may also be found by the method of § 134. The equa- 
tions of the tangents then are 



x cos a -f- y sin a = \Za 2 cos 2 a + b 2 sin 2 a 



and — # sin a -f y cos a = v(a 2 — X) cos 2 a + (6 2 — X) sin 2 a. 

Then square each and add to eliminate a. 

EXERCISES. 

1. Prove that the square of the semi-diameter of an ellipse 

parallel to a tangent at its intersection with the confocal 
x 2 y 2 

a 2 — X b 2 — \ 

2. Find the confocal hyperbola through a point on the ellipse whose 

eccentric angle is \L>. , x 2 y 2 » ,, 

,1ns. — r— - -t^Vt = a 2 - fe 2 . 
cos 2 yf/ sm 2 ^ 

179. If d x and e? 2 are J_s from the center on parallel tan- 
gents to the confocals 

!*L 4_ t = 1 

a 2 "^ 6 2 

£C II 

and + jjZ— = 1, 

a 2 — A 6- — A 

then dj 2 — rf 2 2 = a constant. 

Let the parallel tangents be 

x cos a f i/ sin a = J9, 
# COS a -|- y sin a == p v 



226 ANALYTICAL GEOMETRY 

then, d 2 = [p 2 ] = a 2 cos 2 a -f- b 2 sin 2 a, 

d 2 = [p 2 \ = (a 2 - X) cos 2 a + (b 2 - X) sin 2 a. 
.*. d 2 — d 2 2 = \ (a constant). 

EXERCISES. 

1. Parallel tangents are drawn to a confocal system. Show that the 
locus of the points of contact is a rectangular hyperbola. 

2. TP and TQ are tangents to two confocals and meet at right angles. 
Prove that the line joining T and the center bisects PQ. 

Note. — Points on two ellipses which have the same eccentric angle 
are called corresponding points. Thus, (a cos ^, b sin \}/) and (a x cos \p, 
b v sin \p) are corresponding points on the ellipses 



x< y< 

a 2 ^ b 2 



1, and 4+rC=l. 
a, 2 6,' 



3. [Ivory's theorem.] P and Q are any two points on an ellipse, and 
p, q, the corresponding points on a confocal. Prove Vq = Qp. 

4. Prove that the corresponding points on a system of confocal 
ellipses lie on an hyperbola. 

5. Given the focus and directrix of a conic, show that the polar of a 
given point with respect to it passes through a fixed point. 



X- 



Y 




^ 180. General equation of 


M 


(x,y)PS^ 


a conic. — * 


^^r 

s 1 1 


Let P (x, y) be any point 




/ /I 


on the cnrve. Eccentricity 


, 


( •'! 


= e, F is the focus, YY' the 


L 


V F Q 


A directrix. Take YY' and 
OF J_ to it as axes. Let 
OF = a. 
Now, FP = e • PM = ex, 


Y 


Fig. 131. 


FQ = x - a. 



* Note. — This and the following § should be omitted until the stu- 
dent has read the next chapter. 



CONFOCAL CONICS 227 



v2 



Also, FP* = PQ* + FQ* 

or, e 2 x 2 = if 4- (x — «) 2 . 

... (l _ e 2)2 + 2/2 _ 2 ax + ^ 2 = . . . (1) 

Discussion. — Compare (1) with the general equation of the second 

degree. 

.-. 2 - 1 - e 2 . 
A = a 2 e 2 . 

(1) Now [when the focus is not on the directrix, and A zfz 0], 

(a) If e < 1, 2 > 0, . •. the conic is an ellipse. 

(b) e = 1, S = 0, .-. the conic is a parabola. 

(c) e > 1, 2 < 0, .-. the conic is an hyperbola. 

(2) [When focus lies on directrix, and A = O.J 

(a) Ife<l, 2>0, .-. conic is a point (0, 0). 

(b) e = 1, S = 0, .-.a straight line. 

(c) e > 1, S < 0, . •. two intersecting straight lines. 

Query. — What does equation (1) represent if e= ? 

181. Equation of a conic through given points. — The 

general equation of the second degree, viz., 

Ax 2 + 2 llxy + By 2 + 2 Gs + 2 Fy + C = 0, 
contains five arbitrary constants. 

(1) Hence a conic can be found which shall pass through 
five given points, since each point gives one relation between 
those constants. If, however, three of the given points are 
collinear, the conic reduces to two straight lines. 

(2) If a parabola is required, we have AB — H 2 = 0. 
Hence, only four additional relations are necessary; i.e., a 
parabola can be drawn through any four points, provided no 
three are collinear. 

(3) If a circle, A = B, H = 0, .-.a circle can be drawn 
through any three points not collinear. 

(4) If a pair of straight lines, A = 0, hence they can be 
drawn through any four points. 



A\ \I,Y I ICAL (JKOMHTIiY 



EXERCISES FOR ADVANCED STUDENTS. 

1. Two oonloa have a fooua F Id ooinmon. A variable line through 
K meets the oonloa In P and Q, Prove that the loous of the Interaeotlon 
of tangents at P and <i is a straight line. 

2. Tangents at A, i>, C, to a parabola //•' I u.r [focus at F], form a 
triangle. U R Is the radius *>r Its olrouinsoribed oirole, 

... FA FB • FC 

Trove, |{ — . 

•I a 

3. Kind t iio Loous of the Intersection o{ i tangents to the oonfooal 
parabolas 

y* * I a, (x I a,), 
//• l.i, (.r I a..). 

Aiis. The straight line, x I a, I o a — 0. 

4. show that the equation, 

<• tan'-' a r (see a | cos 0), 

where a Is b variable parameter, represents a system of oonfooals. 

6. if a point, move on the direotor oirole of the ellipse ., | ,.j™ 1» 

its polar envelops, /. <•., always tOUOheS, the confocal. 
■ r ' , t ■ 1 



LV«' J I lr] | \ it- I /> J 



6. Given a system of oonfooals, tool F, F . A fixed straight line 
passes through the loeus F, outting the oonfooals. show thai the tan- 
gents to these various oonios, at their points i^ Intersection with the 
tixed seoant, are also tangents to the parabola desoribed with F as foous, 

anil the secant as directrix. Also, the portion of each tangent Included 
between the eouie ami the parabola, subtends a constant angle at the 
focus F . 

7. Prove that the line joining the points of oontaot of two I tangents 
to the oonfooals whose semi-axes are (a,, b x ) and w.,/>.) respectively, 

envelops the confocal whose semi axes :uv 



LvV + 6, a ' Va, a + bA 



8. show that an equilateral hyperbola, confocal to an ellipse, inter- 
cepts on the sides of a right angle oiroumsoribed about the el i ipso, two 

chords whioh are equal 



CONFOCAL CONICS 229 

9. Given two confocal ellipses. From any point P tangents are drawn 
to one of them, and meeting the other, one in A, B, the second in C, I), 

0V6 ' 1'A^PB'PC PD' 

10. Find the condition that the two conies, 

A lX 7 + 2 U x xy + B, y 2 - 1, 
A 2 z*+2ll 2 xy + B 2 y> = 1, 

may be placed so as to be confocal. 

(A, - B,)2+4II,' = ( A 2 - B 2 ) 2 + 4 H 2 ^ 
(A^-ll^ (A 2 B 2 -H 2 ^ 



CHAPTER XI 
THE GENERAL EQUATION OF THE SECOND DEGREE 

182. We have seen that the equations of the conic sections 
and the circle are all of the second degree, whether the axes 
of reference be rectangular or oblique. In this chapter we 
shall show that the general equation of the second degree 
represents a conic or some special case, by transforming co-» 
ordinates, and reducing it to one of the special forms, wherein 
its locus becomes easily recognizable. 

Let the general equation be written, 

Ax 2 + 2Hxi/ + Bif+ 2Gx + 2Tij + C = . . (1) 

The condition that (1) should represent a pair of straight 
lines was found in [§ 48], viz., A = 0. 

183. Condition for a central conic. — If (1) represents a 
central conic, the terms in x and y should vanish when the 
origin is transferred to the center. For if (x, y) is a point on 
the curve [referred to its center], then (— x, — y) is also a 
point on the curve. But only equations of the second degree 
can be satisfied by equal and opposite values of (x, y). Hence, 
the transformed equation cannot contain any terms in x and y. 

184. Transformation to center. — Let (x v y t ) be the cen- 
ter. Then, transforming to parallel axes through (x lf y), 
equation (1) becomes, 

A(x + xtf + 2H(z + Xl )(y + y t ) + B (y + ytf+2G(x + x,) 
+ 2E(2/-f 2/0 + = 0, 

230 



GENERAL EQUATION OF THE SECOND DEGREE 231 

or, Ax 2 + 2 Kxy + Bif + 2 x (Ax, + By t + G) 

+ 2 2/(11^ + 1^ - F F) + C 1 = . . (2) 

[where C, = Ax, 2 + 2 Hx l2/l + B^ 2 + 2 Gx,+ 2 ¥y t + C] (3) 
The new origin will .-. be at the center if 
Ax, + Hy, + G = 



HF - BG GH - AF 

whence, s I= AB _ H2 > ^= AB-H 2 ' 

Hence the curve has a center if AB — H 2 7^ 0. This ex- 
pression is usually denoted by the symbol 2 (sigma). If 
5 = 0, the center is infinitely distant from the origin ; i.e., 
there is no center. 

Also, from (3) the value of C, may be written, 

C, = x, (Ax, + Ry, + G) + y t (Hx, + B Vl + F) + G^ + F 1/l + C, 
or C t = Gx t + F?/ t -f- C [from equations (4)]. 

HF - BG GH-AF; 

•• C i = G ' AB - H 2 + F ' AB - H 2 + ° 

ABC + 2 FGH - AF 2 - BG 2 - CH 2 



AB - H 2 
* A 

5' 
.-. equation (1) [i.e., the conic referred to its center but not 
to its own axes] becomes, 

Ax 2 + 2 Hxy + B?/ 2 + ^ = . . . . (5) 

185. Equation of conic referred to its own axes. — To 

find this, we must revolve the present axes through an angle 0, 
such that the term in xy of equation (5) will vanish. Now 

put 

x cos — y sin 9, f or x ) . , . /r , N 

_ a , /» * f m equation (5). 

and a; sin H t/ cos 0, tor ?/ ) w 



232 ANALYTICAL GEOMETRY 

The result is, Lx 2 + Mxy + % 2 + ^ = (6) 

where L = A cos 2 + B sin 2 + 2 H sin (9 cos . . (7) 
N" = A sin 2 + B cos 2 - 2 H sin (9 cos . . (8) 
M = 2 (B - A) sin cos + 2 H (cos 2 - sin 2 0) (9) 

Now put M = 0, .-. (A - B) sin 2 - 2 H cos 2 = 0. 

•■• tan2 * = iS ( 10 > 

Hence, if the axes of reference for equation (5) be turned 
through an angle 0, determined by the equation 

O TT 

tan 20= — , 

A — B 

the term in xy will disappear, and the equation of the curve 

referred to its own axes through the center becomes 

Lz 2 +% 2 +^ = (11) 

from which the curve may be constructed. .*. the general 
equation of the second degree represents a conic. 

L and N can be determined in terms of A, B, H. Thus, 
from (7) and (8), 

L + N = A + B, 

L - N = (A - B) cos 2 + 2 H sin 2 0, 

which may be further reduced, finally yielding the result, 
L • N = AB - H 2 = X 

Hence, the curve is an ellipse or hyperbola according as L and 

A 
N have like or unlike signs, provided — is positive. This work 

-^ 

is, however, unnecessary to the construction of the locus of 
equation (11). 

Note. — Observe that the equations found at the various steps of the 
transformation are also of the second degree. 



GENERAL EQUATION OF THE SECOND DEGREE 233 

186. Determination of the kind of conic. — In the gen- 
eral equation, 

Ax 2 + Bif + 2 Kxy + .2 Gx + 2 % + C = 0, 
put p cos for x, p sin for y, obtaining 

p 2 J A cos 2 + 2 H cos sin + B sin 2 0J 

+ 2 P JG cos ■+■ F sin j + C = 0. 

One root of this quadratic in p is infinite if the coefficient 
of p 2 = 0. 

.-. , A cos 2 6 + 2 H sin cos 6> + B sin 2 0=0, 

or, A + 2 H tan 6 + B tan 2 = 0, 

which is the equation determining the lines through the origin 

meeting the curve at infinity ; or, putting — for tan 0, we get, 

Ax 2 + 2 Hzs/ + Bt/ 2 = 0. 

These lines are real if H 2 — AB>0, hence in this case the 
general equation represents an hyperbola, and the lines are 
parallel to the asymptotes. 

They are imaginary if H 2 — AB < 0; hence in this case the 
curve is an ellipse. 

They coincide if H 2 — AB = ; hence in this case the 
curve is a parabola. 

Both roots of the above quadratic are infinite if also G = 0, 
and F = 0. 

Hence, the asymptotes of the conic 

Ax 2 + 2JIxy + Btf+C = 0, 
are Ar 2 + 2 Kxy + By 2 = 0. 

Note. — The asymptotes of an ellipse are imaginary, viz., 

X - + t = 0, 
a 2 + 6 2 ' 

a b 



234 ANALYTICAL GEOMETRY 

187. Condition for a rectangular hyperbola. — The equa- 

tion Ax 2 + 2 Kxij 4- By 2 = 

represents, as we have seen, a pair of straight lines parallel 
to the asymptotes of the general equation. Now, if the latter 
be a rectangular hyperbola, the asymptotes, and .\ the two 
lines given, are _L to each other ; hence A + B = 0. 

[See § 46]. 

188. The asymptotes of a conic. — The equation 
Ax 2 + 2 Hxy + By 2 + 2 Gx + 2 Fy + C + X = 

represents the asymptotes of the general equation if 

A H G 

H B F =0, 

G F C + A 

or A • % + A = 0. 

Hence, if S = denote the general equation, its asymptotes 
are A 

And since the conjugate hyperbola is S + 2 A = 0, its asymp- 
totes are 2 A 

189. The axes of a conic. — The equation, 
tan 2 6 



A-B' 

may be written thus : 

H (1 - tan 2 0) = (A - B) tan 0; 
or, putting - for tan 0, we get, 

H (x 2 - y 2 ) = (A - B) xy, 
which is evidently the equation of the axes of the conic. 



GENERAL EQUATION OF THE SECOND DEGREE 235 

190. Tangent, diameter, and polar at (x lf y^. — These are 
worked out in precisely the same manner as in previous 
chapters, the results being : 

(1) A tangent at (x lf y x ) is 

Axx x + K(x yi + x x y) + By Vl + G (x + zJ+V (V+Vd + C = °- 

(2) A diameter is 

Ax + Hz/ H- G + m (Hz + By + F) = 0, 
where ra is the slope of the given chord. 

(3) The polar of (x 1 , y^) is the same as the equation of the 
tangent. 

Note. — (1), (2), and (3) hold whether the axes be rectangular or 
oblique. Also, the remarks concerning polars in previous chapters are 
applicable here. 

191. Conjugate diameters. — If the above diameter be par- 
allel to y = m v x, we have, 

A + Hra 

or, Bmm l + H (ra + ra x ) + A = 

is the condition existing if y = mx and y = m x x are parallel 
to two conjugate diameters. 

Note. — The polar of the origin is 

Gx + Fy + C = 0. 

192. Condition that two given lines may be conjugate 
diameters of a given conic. — Let the given lines be, 

ax 2 -f 2 hxy + by 2 = 0, 

and the given conic Ax 2 + 2 Hxy -+- By 2 + C = 0. 

Suppose the given lines be written, 

b (y — mx) (y — ra x x) = (1) 



236 



ANALYTICAL GEOMETRY 



2h , a 

.*. m -f- m. = 7-, and mm. = T • 

o o 

Now, the lines (1) are conjugate diameters of the given conic 

lf Bmm t + H (m + m x ) + A = 

,B«-2h£ + A = 0, 

or the required condition is 

Kb + Ba - 2 Hft = 0. 

193. Summary of the discussion of the general equation 
of the second degree. — We here recapitulate the loci repre- 
sented by the general equation, for the various conditions 
among its coefficients. Three typical examples are worked 
out briefly to show the method of procedure in constructing 
the locus. 



2 


A 


Locus. 


>o 


^o 


An ellipse. [A circle when A = B and 
H = O.] 




>o 


Imaginary. 




<o 


Real. 




= 


A pair of imaginary straight lines or a 
point. 


= 


7^0 


A parabola. 




= 


Two parallel straight lines or one straight 
line, or no locus. 


<o 


^o 


An hyperbola. 

If A + B = 0, a rectangular hyperbola. 




= 


A pair of real intersecting straight lines. 



Note. — Several minor details which the student can discover for him- 
self in working examples, are not given here. 



GENERAL EQUATION OF THE SECOND DEGREE 237 



Example 1. Trace the conic. 

hx 2 + 6xy + 5if -16 x-16y+ 8 = 0. 
Here 2 = 16, A<0, .-. an ellipse. 

Center is found to be the point (1, 1). The equation referred to 
parallel axes through the center is 

5 x 2 + 6 xy + 5 y 2 - 8 = 0, 
2H 6 



tan 2 



45°. 



A- B 

Turning axes through 45°, the equation of the ellipse referred to its 



own axes is 



or, 



x 2 + 4 y 2 = 4, 
= 1. 



g 2 | y 2 
(*) 2 (i) 2 



.*. semi-axes are 2 and 1. 
The curve is shown in the figure. 
Y 




Fig. 132. 
Ex. 2. Trace the locus 

4 x 2 + 4 xy + if - 18 x + 26 y 4- 64 = 0. 
Here 2 = 0, A ^ 0, .-. a parabola. 

2 H 4 2 tan 

tan 2 = — = - •-= — . 

A - B 3 1 - tan 2 6 

.-. tan = - 2. 

Hence, if the axes are turned through tan— i (—2), the new equation 

will be, 



V 5/ 4V5\ 10/ 

Hence vertex is point [ — , — -), 
\10 5/ 



latus rectum =4 



14 



4V3 v 7 5 



238 



ANALYTICAL GEOMETRY 



Similarly, the focus is found to be the point (2, — 3), and the directrix 
the line x — 2 y = 1. 

The curve is shown in the figure. 

V 




Fig. 133. 



Ex. 3. Trace the conic 

4 x 2 _ 24 xy + 11 y 2 + 40 .r + 30 y - 105 = 0. 
2 = AB - H 2 = 44 - (- 12) 2 = - 100, A ■£ 0, 



r 


6 


Y' / 






•v 










Fig. 134. 



GENERAL EQUATION OF THE SECOND DEGREE 239 

3 

4 \ 
-24 24 



.'. an hyperbola. .-. tan 6 



tan 2 



B -7 



The center is (4, 3), and the equation referred to parallel axes through 
the center is 

4 x 2 - 24 xy + 11 y 2 + 20 = 0. 

Turning the axes through tan~i - , we get the equation of the hyper- 
bola referred to its own axes, viz., 

t _ t.-i 

4 1 

Hence, semi-axes are 2 and 1. 
The curve is shown in the figure. 

EXERCISES ON CHAPTER XL 

Trace the following loci: 

1. 4 (x — y) 2 = 4 (x + y) — 1. Ans. Parabola. 

2. x 2 + xy + y 2 - 3 x - 3 y = 0. 

.4ns. Ellipse, = 135°, semi-axes y/6, v2. 

3. 2 rr 2 — 3 .r — 4 y — 5 = 0. Ans. Parabola. 

4. 6 re 2 — xy — y 2 — x + 3 y + 2 = 0. Ans. Hyperbola. 

5. ?/ 2 — 3 a; — 4 y = 0. .4 ns. Parabola. 

6. 5 x 2 + 11 xy + 2 y 2 - 13 .r + 10 y - 28 = 0. 

^4ns. Two intersecting lines. 

7. (x - ?/) 2 - 4 - x 2 = 0. 

8. 2x 2 - 2xy + y 2 - 2y = 0. Ans. Ellipse. 

9. 5 .t 2 + 4 at/ + 8 y/ 2 - 18 x - 36 y + 9 = 0. 

Ans. Ellipse, = tan _1 ( — £), semi-axes 3 and 2. 

10. (x - 2 t/) 2 - 2 (.r + 2 y) + 1 = 0. Ans. Parabola. 

11. xy — bx — a?/ = 0. 

Ans. Hyperbola, 6 = 45°, semi-axes (V2 ab, \ / 2~ab). 

12. (3 x + 4 ?/) 2 + 22 .r -f 46 y + 9 = 0. 

Ans. Parabola, latus rectum = |. 

13. 6x 2 — xy — y 2 —x + 3?/ — 2 = 0. Ans. Two straight lines. 

14. 4x 2 — 4.x?/ + y 2 + 8.c — 4?/ = 0. Ans. Two parallel lines. 



240 ANALYTICAL GEOMETRY 

15. Sx 2 + 2y 2 - 2x + y - 1 = 0. Ans. Ellipse. 

16. x 2 + 2 xy — y 2 + 8 x + 4y — 8 = 0. Ans. Hyperbola. 

17. Vx + Vy = Va V8. 

Ans. Parabola referred to tangents at extremities of latus rectum [4 a]. 

y 4a 2 6 2 



'•vW ; 



18. K/'- 4- \/v =!• Ans. Parabola, latus rectum 



6 ' [a 2 + b 2 ]! 

19. Find the center of the conic 

Ax* + 2Hxy +By 2 - 2y =0. 

AnS ' \H 2 - AB' H 2 - AB; 
Trace the conies. 

20. (A 2 + B 2 ) (x 2 + y 2 ) = (Bx + Ay - AB) 2 . 
Ans. Parabola, latus rectum 



2AB 



VA 2 + B 2 

21. xy— x 2 — a?=0. Ans. Hyperbola. Product of semi-axes = 2 a 2 . 

22. \/x + \ // y = 3. Ans. Parabola, latus rectum = 3 V2. 

23. Find the condition that the four points of intersection of the 
conies 

A, a; 2 + 2K l xy + B iV 2 = 1,) 

A 2 z 2 + 2H 2 xi/ + B 2 i/ 2 = 1,J 

may be concylic. Ans. (A 1 — BJ H 2 = (A 2 — B 2 ) H x . 

24. The major axes of two conies are parallel. Show that their foui 
points of intersection are concylic. 

25. Show that the eccentricity of the conic 

Ax 2 + 2 Rxy + By 2 + 2 Gx + 2 Fy + C = 
is determined from the equation 



l_2 - e 2 \ 



[A _ B] 2 + 4 H 2 
[A + B] 2 



26. Show tnat the equation of a conic referred to an axis and the 
tangent at the vertex is 

y 2 = px + qx 2 , 
and is (1) a parabola, if q = 0; 

(2) an hyperbola, if q > 0; 

(3) an ellipse, if o < 0. 

Remark. — The names parabola, ellipse, and hyperbola were originally 
derived from this property. 



CHAPTER XII 
HIGHER PLANE CURVES 

194. Definitions. — A curve whose equation in Cartesian or 
rectilinear co-ordinates involves only algebraic functions is an 
algebraic curve. Any other curve is called a transcendental 
curve. Examples : 

x = cos y, y = sin -1 x, 

y = tan -1 x, x = log a y, etc. 

Transcendental curves and all algebraic curves whose equa- 
tions are of a higher degree than the second are known as 
higher plane curves. A few of these, which have become 
celebrated in the history of mathematics, will be examined 
in the present chapter. 

195. The cissoid.* Definition. — AL is a tangent to the 
circle OFA at A, the end of the diameter OA. C is the cen- 
ter of the O. OB is any straight line from to AL and cut- 
ting the circle in D. Take OP = BD. The locus of P as OB 
revolves about 0, is the cissoid. To find its equation, take 
OX and OY _L to it as axes. Let P be (x, y) and the radius 
CA of the circle equal to a. 



* Invented by Diocles, a Greek geometer of the second century b. c, 
for finding two mean proportionals between two given lines, or more 
specifically to "duplicate the cube 11 ; i.e., to find the side of a cube 
whose volume shall be twice that of a given cube. 

241 



242 



ANALYTICAL GEOMETRY 






Fig. 135. 


Now, 


MP : OM : : ND : ON ; 


also, 


OM = NA [since OP = BD], 


and 


ON = 2a - x, ND = V(2a - x)x. 




. V _ V(2 a — x) x 2 x 3 




x 2a -x ; 0f V 2a -x 



is the equation required. 

Discussion. — (1) Curve is symmetrical to z-axis. 

(2) It lies entirely between the lines x = 0, x = 2 a. 

(3) It passes through the extremities of the diameter parallel to the 
i/-axis. 

(4) It has two infinite branches to which the line x = 2 a is an 
asymptote. 

To find the polar equation, take for the pole and OX for the initial 
line. Let P be (/a, d). Then, 

p = OP = DB = OB - OD. 
= 2 a sec 6 — 2 a cos 0. 
= 2« (sec 6 — cos 6). 
.'. p = 2 a sin 6 tan 0, 
which is the equation required. 

The discussion is left to the student. 



HIGHER PLANE CURVES 



243 



196. To duplicate the cube. — Take CK = 2 a. Draw AK, 

cutting the cissoid in H. Now, since 

CA = i CK, GA = i GH. 
But from the equation of the cissoid, we have, 
OG 3 



GH = GA 

Now, let c be the 
required such that c* = 

or i 

But 



JGH 



GET = 2 OG*. 



edge of any given cube, and let c x be 
= 2 c 3 . To this end, take c x so that 
OG : GH : : c : c x 
JG^GH 8 :^: q 8 . 
GH 3 = 2 0G 3 . 

■ ^3 = 9^ 



Hence, c t is the edge of a cube whose volume is double that 
of the given cube whose edge is c. In like manner, by taking 
CK = ka, we can find the edge of a cube k times the volume 
of a given cube. 

197. The conchoid.* Definition. — Let A be a fixed point, 
Y 




Fig. 136. 



* Invented by Nicomedes, a Greek geometer of the second century 
b.c, for the purpose of duplicating the cube. It is, however, more 
easily applicable to the trisection of a given angle, which was also a 
famous problem of the ancients. 



244 



ANALYTICAL GEOMETRY 



and OX a fixed line. A straight line LL' revolves about A, 
cutting OX in Q. The locus of the point P such that PQ 
always equals a given constant, a, is the conchoid. A is the 
pole, OX the directrix, and the constant distance PQ [a] is 
the parameter. To rind the equation of the curve, take OX. 
and OY _L to it through A, as axes. 

Draw AN II to OX and meeting the ordinate PM in 1ST. 

Let P be (x, y). Let AO = c [a known distance]. 

Then by similar As, 

AN : NP : : QM : M P, 
or, x : y + c : : Va 2 — y 2 : y, 

whence, x 2 y 2 = (y + c) 2 [a 2 — y 2 ~\ 

is the required equation. 

Discussion. — (1) The curve lies between the lines y = -±- a. 

(2) It is symmetrical to the ?/-axis. 

(3) The curve has four infinite branches to which the z-axis is an 
asymptote. 

(4) If c < a, the curve has an oval or loop beneath A. 
If c = a, the loop becomes a point at A. 

If c > a, both portions of the curve lie above A. 
If c = o, the conchoid becomes a circle. 
To find the polar equation of the curve, let A be the pole, AY the 
initial line, and P or P x (p, 6). 
Then, p = AP = AQ -|- QP 

= OA sec 6 -J- QP. 
.-. p = c sec 0-J- a 
is the equation required. 

198. To trisect a given angle. — Let BAG be the given 

angle. On AB lay off AL 

[any length], and through 
L draw LO _L to AC. Now 
take OC = 2 AL, and with 
A as pole, OX, directrix, 
and OC as parameter, con* 
Struct a conchoid, 



L 


X' 





L 


^ 






f/f w, 



HIGHER PLANE CURVES 



245 



Draw LD J_ to OX. Then AD cuts off a third of the given 
angle BAC. 

Proof : Bisect DF in E. Then 

LE = ED = i OC = AL. 
.'. Z LAE = Z LEA = 2 Z LDE = 2/ DAC. 



. Z DAC = 



i Z BAC. 



199. The lemniscate. — This curve is the locus of the foot 
of a perpendicular from the center of a rectangular hyperbola 
to a tangent to the latter curve. 

To find its equation : 




Fig. 138. 

The tangent to the hyperbola x 2 — y 2 = a 2 at (x lf y t ) is 

xx 1 — yy 1 = a z (l) 

The _L to it from the origin is 

x 1 
or, X -=- % L (2) 

Also, x 2 — y 2 = a 2 (3) 

Eliminating (x v y t ) between (1), (2), and (3), we get, 

(x 2 + ?/V = a 2 (z 2 -2/ 2 ), 
which is the equation required. 



24G 



ANALYTICAL GEOMETRY 



Discussion. — (1) The curve is symmetrical to both axes, and .-. to 
the center of the hyperbola. 

(2) It lies entirely between the lines x = J- a. 

(3) It passes through the origin or center. Its properties can be 
more readily seen from its polar equation. Put x = p cos 6, y = p sin 6. 
The result is, 

p2 = a 2 cos 2 e. 



p = _J_ a V cos 2 6. 



i a. When d < 45°, p has 
4o°, p = 0, hence the curve 



Discussion. — When 6 = 0° or 180°, /o 
two equal opposite values < a. When 6 
cuts the initial line [#-axis] at 45°. 

When 6 lies between 45° and 135°, p is imaginary. When 6 lies be- 
tween 135° and 180°, p has again two equal and opposite values < a. 

Hence the curve consists of two ovals. Also, the asymptotes of the 
hyperbola are tangents to the lemniscate at 0. • 

200. The witch.* Definition. — The line YA is a tangent 
to the circle whose center is C, at the end of the diameter OY. 
Any line OK from O to YA cuts the circle in M. If the ab- 
scissa NM of the point M be produced to P so that NP = YK, 
then the locus of P, as the line OK revolves about O, is the 
witch. 



Y 


K 


A 


y ( n 




\\l y 

ySEs 




^<r c 








v 






X 



o 

Fig. 139. 



To find its equation, take OY and OX _L to it as axes. Let 
OY = 2 a, and let P be (x, y). 



Invented by Donna Maria Agnesi (1718-1799) an Italian mathema- 



tician, 



HIGHER PLANE CURVES 



247 



or 



Then, by similar As, 
ON : OY 
y : 2a 



x 2 y 



NM : YK [= NP] 

\ly(2a- y) : x. 
4a 2 (2a-2/) 



is the required equation. 

Discussion. — (1) Curve is symmetrical to the ?/-axis. 

(2) Lies entirely between the lines y = 0, y = 2 a. 

(3) It has two infinite branches to which the rr-axis is an asymptote. 

201, The cycloid, — This curve is traced by a point P on 
the circumference of a circle which rolls on a fixed line OX. 
This line is called the base, the rolling circle the generatrix, 
and P the generating point. 



H 





^^ * ' ' 


\ ^^^^. 






\ ^^^ 


1 \^r^ 


\ / 


\ ^^ 




\ / 


\ ^^ 


I /^ G 


f 


1 x 




1 \ 


/ X 


J \ 


/ X 


f 1 ^ — 1 _--^ 







OM 



P 

Fig. 140. 



To find its equation : take OX [the base] as z-axis, and the 
point where the curve begins [I.e., the position of P when 
the circle commences to roll] as origin. 

Let the radius of the circle be a and /_ POT = 0, where C 
is the center of the circle. 

Then, arc PT = segment OT, over which the circle has 

rolled, and 

arc PT 
= , .*. arc PT = ad. 



Let P, any point on the curve, be (x, y). 

Now, x = OM = OT - TM = arc PT - PK 



248 ANALYTICAL GEOMETRY 

.-. x = aO — a sin 6 = a (0— sin 0) . . , (1) 

Similarly, y = a (1 — cos 6) (2) 

To eliminate 6, we get from (2), 

cos 6 = *=!, .-. 6 = cos- (±Zl\ = vers-, g) ; 

Also, sin 6 = • 

a 

Substituting these values in (1), we obtain 
x = a vers -1 — — V2 ay — y 2 , 

which, is the required equation of the cycloid. 

Note. — Equations (1) and (2) are also taken together as the equa- 
tions of the cycloid. 

202. The hypocycloid, — This is the locus of a point P on 
the circumference of a circle which rolls on the inside of a 
fixed circle. 




Fig. 141. 



To find its equation : Take 0', the center of the fixed circle, 
as origin, and the _L diameters OAX, OY^, as axes. Let A be 
the initial point of the curve. 

Eadius of fixed circle = a ) np _ 

Radius of moving circle = b ) 



HIGHER PLANE CURVES 



249 



Let P (x, y) be any point on the curve. Draw 0'OT through 
the center of the moving circle. Also the ordinates ON 
*nd PM, and PR parallel to O'A. 

Now, arc TP = arc TA [covered by moving circle]. 

x = O'M = O'N + M = O'N + PR 
= O'O • cos 6 -f- PO • cos <f> = O'O cos + PO cos (6 f - 6). 

.'. x = (a - b) cos 6 + b cos ($'- 0) ; 
but aO = W, 



o 



.'. x = (a — b) cos + b cos ( — — j 
Similarly, y = (a — 6) sin — b sin ( — - — J 6 



(i) 

(2) 



These two equations together represent the hypocycloid. 
A single equation can, however, be obtained by eliminating 6. 

Note. — When the radii are commensurable, the hypocycloid is a 
closed curve. 

203. The hypocycloid of four cusps — This is the hypo- 
cycloid generated when a = 4 b. 

Equations (1) and (2) of § 202 
become, 

x — I a • cos 6 + \ a . • cos 3 . (3) 
y = J a • sin — | a ' sin 3 . (4) 
But, cos 3 = 4 cos 3 0-3 cos 0, 
sin 3 6 =- 3 sin (9-4 sin 3 0. 
.'. (3) and (4) become 
x = a cos 3 0, 
y — a sin 3 9, 

whence jc§ -\- yi = c$ 

is the equation required. 




f/gr. 742. 



250 ANALYTICAL GEOMETRY 

# 

204, The limacon,* — If in the circle whose equation is 
p = 2a cos 0, a constant length be added to each radius vec- 
tor, we get, p = 2 a cos 6 -+- k, which is the equation of the 
limaqon. 



X -o 



Fig. 143. Fig. 144. 

Generally we take k = a. .'. p — a (1 -f 2 cos 6). 

Discussion. — The curve is symmetrical to the initial line. 

The polar equation is easily transformed to rectangular 
co-ordinates by putting 





p = var + ?/ , sm 6 = 
The result is, (x 2 + y 



cos 



Vz 2 + y 2 y/x 2 + y 

-2ax) 2 = a 2 (x 2 + y 2 ). 
2 a, the curve is called the cardioid. 



Its 



2 a (1 + cos 6), 
2 a (1 - cos 6) 



Special Case. — If k = 
equation is, therefore, 

P 
or, p 

[by turning axes through 71-]. 
Its rectangular equation is, 

(x 2 + y 2 + 2 ax) 2 •■= a 2 (z 2 + y 2 ). 

205. The rose of four loops. — This curve is the locus of 
the foot of al from the origin on a line which is of fixed 
length and moves with its extremities always in the axes. 



* Invented by Pascal, a French philosopher (1623-62). See note 10, 
Appendix. 



HIGHER PLANE CURVES 



251 



PQ is of constant length. OM is the _L on PQ. The point 
M describes the rose. The _L OM is a maximum when M 
falls at the mid-point of 
PQ. [For the altitude QJ 

of a right A of a con- 
stant hypothenuse is 
greatest when the A is 
isosceles.] 

To find its equation. 
Let be the pole, OX 
the initial line. Then, 
from the right A OMP 
and OPQ 

p = OP cos 0, OP = 2 a sin 0. 
.-. p = a sin 2 6 

is the equation required. 




Fig. 145. 



[2 a = length of PQ.] 



To obtain its rectangular equation, put 

p = Var + if, sin $ = - , 
P 

The result is, (x 2 + if) 3 - 4 crxhf = 



cos 6 = 



0. 



[The equation is of the sixth degree].* 

Note. — The mid-point of PQ describes a circle of radius a. 



SPIRALS 

206. Definition. — A spiral is the locus of a point which 
revolves about a fixed point or pole, while its radius vector 
and vectorial angle continually increase or continually de- 
crease accordinsf to some law. 



* The discussion of these equations is left to the student. The polar 
equation will more readily show the properties of the curve. 

The equation p = a cos 2 d represents the same curve turned through 
an angle of 45°, i.e., its branches lying on the axes. 



252 



ANALYTICAL GEOMETRY 



The portion of the curve generated in one revolution of the 
radius vector is called a spire. 

The measuring-circle of the spiral is the circle whose center 
is at the pole, and whose radius is equal to the length of the 
radius vector at the end of the first revolution, i.e., the value 
of p when = 2 tt. 

207. The spiral of Archimedes. — This curve is the locus 

of a point whose radius vector bears a constant ratio to the 

vectorial angle. The equation follows from this definition; 

viz., 

P 



k 



Discussion. — The spiral passes through the pole. Also, the radius 
vector increases without limit as the number of revolutions increase cor- 
respondingly. 




Fig. 146- 



e 





TV 
I 


7T 

2 


3 
4" 


77 


5 

I" 


. . . 


etc. 


p 





k- 


k- 
k 2 


k 3 ^ 
* 4 


k • 7T 


k-rTT 

4 




etc. 



HIGHER PLANE CURVES 



253 



From the table, a definite number of spires are readily con- 
structed. The measuring circle here is ABC, whose center is 
and radius OC = 2 kw. Again, the spires, being every- 
where equally distant along the radial lines, are said to be 
" parallel." The figure shows the curve for positive values 
of A 

208. The reciprocal spiral. — In this curve the radius 
vector varies inversely as the vectorial angle i.e., as the re- 
ciprocal of the vectorial angle. Its equation is, .*. 

k 
p = -Q> or pd = k. 

The curve is readily described from the following table : 







1 


1 


3 




5 






6 





t* 


2* 


I" 


7T 


I" 


etc. 








o k. 


m k 


8 k 


n k 


8 k 






P 


00 


& 7T 


4 '2^ 


32^r 


2— - 

2tt 


52^ 


etc. 






Fig. 147. 



Discussion. — The spiral begins at infinity [when d = 0] and winds 
continually around the pole, always approaching nearer and nearer but 



254 



ANALYTICAL GEOMETRY 



never reaching it, or reaching it after an infinite number of spires have 
been described, since p = 0, when = <x. 

The circumference of the measuring circle is k. 



[When 



2 7T, p = 



circumference = 2 ir 



-.) 



Again, since p = k, the arc AB described with the radius Vector of 
any point A is constant. The line parallel to the initial line OX and at a 
distance k above it is an asymptote to the infinite branch, for as p ap- 
proaches oo, the arc AB becomes perpendicular to OX. 

The figure shows the curve for positive values of 0. 



209. The parabolic spiral. — In this curve, the square of 
the radius vector varies as the vectorial angle. 

.-. p 2 = k 
is its equation. 




Fig. 148. 

Discussion. —The curve begins at the* pole [when = 0, p = 0] and 
winds continually around it, always receding from it. The radius 
vector becomes infinite after a'j. infinite number of spires have been 
described. 

210. The lituus. — In this curve, the square of the radius 
vector varies inversely as the vectorial angle. Its equation 
is- -• k 



is 



p =~e 



HIGHER PLANE CURVES 



255 




Fig. 149. 
Discussion. — The curve begins at infinity and winds steadily around 



the pole but never attains it. 



Ve = o, p = oo.l 

\_e = oo, p = o.J 



The figure shows the curve for positive values of 0. 

211. General equation of a spiral. — This may obviously 
be written, p = k • n (1) 

Discussion. — When n = 1, equation (1) is the spiral of Archimedes, 
n = — 1, the reciprocal spiral. 
n = \, the parabolic spiral. 
n = — ^, the lituus 

212. The logarithmic spiral. — In this curve, the loga- 
rithms of the radii-vectores are in the same ratio as the 
vectorial angles. 




Fig. 150. 



256 



ANALYTICAL GEOMETRY 



Its equation is, .-. log p = k 6. 

It may also be written in these forms; viz., 
p = a 9 , = log a p. 






-2 


-1 





1 


2 


3 








etc. 












p 


1 
4 


1 
2 


1 


2 


4 


8 








etc. 











When a = 2, the table given here is readily found. [6 is 
expressed in radians.] 



Discussion. — When d 
6 



0. 



— oo, p 

0, p = 1. 

.-. p varies from to + 1, when 6 varies from 
from + 1 to + oo, when varies from to + oo. 
,'. The number of spires is infinite. 



oo to 0, and p varies 



EXERCISES ON CHAPTER XII. 

1. Find the locus of a point the product of whose distances from two 
fixed points is constant. Take the distance between the points equal to 



a V2 and the constant product is — 



Ans. The lemniscate. 



2. Show that the cissoid is the locus of the foot of a perpendicular 
from the vertex of a parabola on a tangent to the latter. The parabola 
is y 2 = — 8 ax. 

3. Show that the cardioid is the locus of a point on the circumference 
of a circle which rolls on an equal fixed circle. 

4. Show that the logarithmic spiral may be defined as the locus of a 
point such that its radius vector increases in a geometric ratio, while the 
vectorial angle increases in an arithmetic ratio. 

5. Construct the logarithmic curve y = a x or x = log a y. Show that 
every logarithmic curve passes through the point (0, 1) and has the 
#-axis for an asymptote. 



REVIEW EXAMPLES 257 

EXERCISES FOR ADVANCED oTUDENTS. 

1. A variable circle touches a given fixed circle in a given point. A 
tangent is drawn common to the two circles. Find the locus of its point 
of contact with the variable circle. Ans. A cissoid. 

2. A variable triangle whose vertex A is fixed, and /_ A is constant, 
is inscribed in a circle. Find the locus of the centers of the circles in- 
scribed in, and circumscribed about, the triangle. Ans. Two limacons. 

3. In a triangle the base a is fixed, and the other two sides b and c, 
and the median m to the base, always satisfy this^relation, viz., 

b - c = m \/2. 
Find the locus of the vertex of the triangle. Ans. A lemniscate. 

4. Two circles are given and a tangent to each. These tangents meet 
at an angle which is constant [and given]. Find the locus of its vertex. 

Ans. Two liinacons. 

REVIEW EXAMPLES LN PLANE ANALYTICAL 
GEOMETRY. 

1. If A, B, C are three points on a straight line, let a, 6, c be their 
distances from a fixed point on the line. 

Prove : (b- a) + (c - b) + (a - c) = 0. 

Suggestion : AB + BC + CA = 0. 

2. Let a, 6, c, d, be the distances of any four points A, B, C, D from 
a fixed point 0, all being on the same straight line. 

Prove : AD • CB + BD • AC + CD • BA = 0. 

Suggestion : 

(d - a) (b -c) + (d- b) (c- a) + (d - c) (a - b) = 0. 
If the points are situated in the order A, B, C, D, we have 

CB = - BC, BA = - AB, 

whence we obtain, 

BD • AC = AD • BC + CD • AB. 

3. Find the distance between the points (2, - 3) and (— 3, 4). 

Ans. n/74. 

4. Find a point equidistant from the three points (8, 6), (2, — 2), 
(8, — 2). Let its co-ordinates be (x, ?/) ; then we have 

(x - 8) 2 +(y- 6) 2 - (* _ 2) 2 + (y + 2)' « (* - 8) 2 + (y + 2) 2 
or -4x + 42/ + 8=-16a:4-4y + 68^^J6a?-12y + 100, 

Solving these equations, we have x ~ 5, y = 2 S 



258 ANALYTICAL GEOMETRY 

5. Express that the point (x, y) is equidistant from (1, 2) and (3, 4). 

Ans. x + y — 5 = 0. 
From Elementary Geometry we know that (x, y) must lie on a line 
which bisects perpendicularly the line joining the given points. 

6. Eind the distance between (1, — 3) and (0, — 2). Ans. V2. 

7. Find the areas of these triangles : 

(a) (-4, -1), (3, -2), (2,1). Ans. 10. 

(jS) (4, - 5), (- 3, - 6), (2, 3). Ans. 29. 

(7) (4, 5), (16, 9), (- 2, 3). Ans. 0. 
i.e. the points are collinear. 

8. Eind the distance between the points of intersection of 

rr 2 + ?/ 2 = 25, &x-Sy = 7. Ans. Iff. 

9. Show that the points (0, 0), (4, — ^), and (4, — j form an equi 



a'. 

Ans. p 2 cos 2 6 = a 2 . 

11. G is the centroid of A ABC. Prove by co-ordinates that 

A GBC = i A ABC. 

12. Eind the equations of lines through the origin and inclined re- 
spectively at 45°, 60°, and 120° to the x-axis. 

Ans. y = x, y = x V3, y = — x Va. 

13. Eind the equation of a line through (1, 2) and at 60° to z-axis. 

Ans. x VS - y = VS - 2. 

14. Eind the line through (0, — 3) and at 45° to z-axis. 

Ans. y = x — 3. 

15. Eind the line through (1, 2) and parallel to the line y = 2 x + 3. 

Ans. y = 2x. 

16. Eind the line through (3, 5) whose intercepts are equal. 

Ans. x + y = 8. 

17. Eind the line through (3, 3) which forms with the axes a triangle 
whose area is 18. Ans. x + y = 6. 

18. Show that the lines 

- + r = 1, r + - = 1, x = y< 

a b b a 



are concurrent m 



the point [ =- » 7 )• 

* \a + b a + bj 



/ 

REVIEW EXAMPLES 259 

19. Find area of A whose sides are 

x - y = 0, x + 3 y = 0, a; + ?/ + 4 = 0. ^ ns< 8. 

Also the A whose sides are 

'Sx+y- 7 = 0, z + 7y+ 11 = 0, a; -3# 4- 1 = 0. ^^s. 10. 

20. Find the equations of the medians of the A whose vertices are 
(2,1), (3, -2), (-4,-1). 

Ans. x - y - 1 = 0, x + 2 y + 1 = 0, x - 13 y - 9 = 0. 

21. Find the equation of the line joining the points (a, b) and (6, a). 

Ans. x + y = a + b. 
Also the line joining the points (a, 6), (— a, — 6). Ans. bx — ay = 0. 

22. Show that these points are collinear ; (1, — 1), (2, 1), (— 3, — 9). 
Find the ratio of the segments into which these points divide the line. 

Ans. \. 

23. Find the equations of (1) The line joining the origin to the 
intersection of y = 2 x — 3 and x + 3 y + 4 = 0. (2) The line j oining 
(—4, 7) to the intersection of y — Sx — 1, x — y — 1. 

Ans. (1) 11 x + 5 y = 0. (2) a: + ?/ - 3 = 0. 

24. Find the lines through the intersection of y = Sx — 1, x = y — 1 
and parallel respectively to the x-axis and 4rc + 5y + 6 = 0. 

Ans. y = 2, 4 x + 5 y — 14 = 0. 

25. Find the line joining the intersection of 

2.r+3y + 3 = 0, .r + 2?/ + 2 = 0, 
to that of 2 x + y + 3 = 0, 3x + 2?/ + 4 = 0. 

Ans. x + y + 1 = 0. 

26. Through the vertices of the A whose sides are 

x + 2 y - 5 = 0, 2.r + y-7=0, y - x - 1 = 0, 
parallels are drawn to the opposite sides. Find their equations. 

Ans. x — y — 2 = 0, x + 2y -8 = 0, 2x + y - 4 =0. 

27. Two lines can he drawn through (6, — 4) each of which forms 
with the axes a A whose area is I ; find their equations. 

Ans. Sx + 4y - 2 = 0, 16 x + 27 y + 12 = 0. 

28. Find the angles between these pairs of lines : — 
(a) 5 y - 3 x + 1 = 0, </ - 4 .r + 2 = 0. 
(|8) 2 .r - 3 ?/ = 0, 6 x 4- 4 ?/ + 7 = 0. 
(7) 2/ + * = 0, (2 + V3) y - x = 0. 

(5) y - kx = 0, (1 -- A') t/ - (1 + A") x = 0. 

Ans. 45°, 90°, 60°, 45°. 



260 ANALYTICAL GEOMETRY 

29. Find the acute angle between the lines 

2z + 3?/-4 = 0, 4z-5?/-6 = 



a .22 

Ans. si 



V533 

30. Find a line through (0, — 1) and ±to x + y — 1=0. 

Ans. x — y — 1 = 0. 

31. Find the lines through the origin and inclined at 45° to 

y + xV~3 = 0. Ans. y + x (2 - x / '3) = 0. 

y - x (2 +V3)=0; 

32. Find the line through (1, 2) and ± to 3^ + 4^+ 5 = 0. 

,4 ns. 4rr - Sy + 2 = 0. 

33. Find the line through (2, 3) and JL to the line joining (1, 2) and 
( - 3, - 14). Ans. x + 4 y - 14 = 0. 

34. Find the distance from (2, 3) to the line 3x + 4y — 20 = 0. 

Ans. — f . 
Also from the origin to 3 x + 4 y + 20 = 0. Ans. 4. 

35. Find the distances from (1, 3), (2, — 3), (0, 0), on the line 

4x-5w + 6 = 0. , 5 29 6 

Ans. —=, —=, —=. 
\/41 V41 Vil 

36. Find the equations of the altitudes of the A (— 4, — 6), (12, 2), 
(3,8). Ans. 2 x + y = 14, x+2y = 16, Sx-2y = 0. 

Find the ortho-center of this A. Ans. (4, 6). 

Also the lengths of these altitudes. 21 24 66 

Ans. —=i — —i — -= 
V5 Vb Vl3 

37. Find the equations of the ± bisectors of the sides of the preced-. 
ing A. .4ns. 2z + ;?/-6 = 0, 2x + 4?/-3 = 0, 6z-4i/-25 = 0. 

Find their point of concurrence. Ans. (3|, — 1). 

38. Find the line joining the feet of the ±s from origin on the lines 
3 x - 4 y + 25 = 0, 12 x + 5 y - 169 = 0. Ans. x - 15 y + 63 =0. 
Find also the distance between the feet of the ±s. Ans. V226. 

39. Find the points on the line Sx — y — 1 = at a distance 2 from 
the line 12 x - 5 y + 24 = 0. Ans. (1, 2), (- 5 /, 64). 

40. Find the center of the inscribed of the A, (1) whose sides are 

3x-4y = 0, Sx + 15?/ = 0, re = 20. 
(2) whose vertices are (2, 5), (1, 2), (3, 4). Ans. (13, 1), (V6, 4). 



REVIEW EXAMPLES 2G1 

41. If the angle between the axes is 60°, find the ± from (1, 1) on 

2x + Sy + 4 = 0. Ans. 4 x - y - 3 = 0. 

Find also its length. 2 V3 

Ans. — — • 

V7 

42. ABC is a right-A, /_ C being 90°. Squares ACED, BCGH, are 
described on AC, BC. Show that BD and AH meet on the ± from C 
on AB. 

43. OIN and OLM are given straight lines ; I and N are fixed points ; 

IL and NM meet at P. Find the locus of P if 

a 8 

Of + (if = *' wnere a and i 3 are constants. 

Ans. A straight line. 

44. Determine k so that 

x 2 + kxy - 8 y 2 + 12 y - 4 = 0. 
may represent a line-pair. Ans. k = ^ 2. 

45. Find the line-pair which joins the origin to the intersection of 

2 x + y = 6, and z 2 + ?/ 2 = a; + 5 y + 6. 

Ans. x 2 + y 2 = x + 5 y f ^— ^ 1 + 6 f — 2-1 

or xy = 0, i.e. the axes. 

Hence the given curves intersect on the axes. 

46. Interpret the following equations : 

(1) (s-8)(y-4)-0. (4) x 2 - y 2 = 0. 

(2) x 2 -?>ax + 2a 2 = 0. (5) xy - 2x + Sy - 6 =0. 

(3) xy = 0. (6) 6xy+2bx + 3ay + ab = 0. 

47. Find the line joining the origin to the intersection of 

- + £ = l and r -f £ = 1. Ans. x = y. 

a b b a 

48. Find the diagonals of the <c whose sides are 

8x-2y = l, 4.r-5?/ = 6, 3x-2y = 2, 4x-5y = S. 

Ans. 13z - 11?/ - 9 =0, 6x-y = 0. 
Find also the area of the ^ 7 . Ans. f. 

49. Show that the lines 

B.r 2 - 2 Hz?/ + Ay 2 = 
are ± to the lines Az 2 + 2 Hzi/ + B?/ 2 = 0. 



262 ANALYTICAL GEOMETRY 

50. Find the condition that one of the lines 

Ax 2 + 2 Bxy + By 2 = 
may coincide with one of the lines 

A^ 2 + 2 R lX y + B^ 2 = 0. 
Ans. 4 (AH, - A,H) (RB 1 - H X B) = (AB X - AjB) 2 . 

51. Find also the condition that a line of the first pair may be ± to a 
line of the second pair. 

Suggestion. — In the preceding condition put B, — H, and A for 
A, H, and B (by Ex. 49). 

52. Find the lines through (a, j8) and ± to the lines 

Ax 3 + Bxhj + Cxij 2 + By 3 = 0. 
Ans. T)(x-a) 3 -C(x-a) 2 (y-p) +B (x-a) (y~P) 2 -A (y-p) 3 = 0. 

53. Find the conditions that two of the lines 

Arc 4 + Bxhj + Cxhf + T>xy 3 + Ey* = 
should bisect the angles between the other two. 

Suggestion. — Put (See §47.) 
Az 4 + Brfy + •••• = 

(A x a: 2 + 2 U,xy + B x y 2 ) (x 2 - - 1 ~ B ' xy - y 2 \ 

Multiply out and equate coefficients. Then eliminate A v H 15 and B v 

Ans. 8A + 3E + C = 0. 
2 (A - E) 2 (A + E) = (B + D) (BE + DA). 

54. Determine X so that two of the lines (See § 46, note.) 

2 xy (Ax + By) — X (x 3 + y 3 ) = may be ±. 
Suggestion : — 

Put • 2xy( )-\( ) = 

(x 2 — kxy — y 2 ) (mx + ny) 
multiply out and equate coefficients. We get equations to determine 
m, n, /c, and X. Ans. X = 0, or X = A + B. 

55. Find the condition that one of the lines 

Ax 3 + 3 Bx 2 y + 3 Cxy 2 + T>y 3 = 
may bisect the /_ between the other two. 

Ans. A (B+D) 3 -D (A + C) 3 =3 (A + C) (B+D) (B 2 + BD-AC -C 2 ). 

56. Any tangent to a circle, radius r, meets the tangents at the 
ends of a diameter AB in P and Q, 

Prove AP • BQ = r 2 . 



REVIEW EXAMPLES 263 

57. Show that these circles are coaxial ; 

x 2 + if + 4 x + y -3 = 0. 
x 2 + y 2 — x — y — 1 = 0. 
x 2 + ?/ +14 x +5^-7 =0. 

58. Find the length of the common chord of the circles 

x 2 + y l - 4 z - 2 y - 20 = 0. 

z 2 + y 2 + 8 x +. 14 2/ = 0. A ns. 8. 

Suggestion. — Find the radical axis. Then find length of chord of 
this axis intercepted by the first circle. 

59. Find the locus of a point which moves so that the tangents from 
it to the circles 

x 2 + y 2 - 3 = 0, .r 2 + y 2 - 3 x - 6 = 0, 
are in the ratio 2 : 3. Ans. The circle 5 (x 2 + y 2 ) + 12 x — 3 = 0. 

60. Find the external common tangents to the circles 

X 2 + yi _ 4 x _ 2 y + 4 = 0, 
rr 2 + ?/ 2 + 4;r + 2?/-4 = 0. 

^Ins. y = 2, 4x—Sy = \0. 
Find also the internal common tangents. Ans. x = 1, 3 x + 4 ?/ = 5. 

61. Find the external common tangents to the circles 

x i + yi _ 6 x - 8 ?/ = 0, x 2 + ?/ 2 - 4 .r - 6 y = 3. 

Ans. x + 2 = 0, y + 1 = 0. 

62. Show that the tangents from the origin to the circle 

(x - a) 2 + (y - b) 2 = r 2 
touch the circle (x - ka) 2 + (y - kb) 2 = (At) 2 . 

Also, if any line through the origin cut these circles in P, Q, P', Q'. 
Prove, OP : OP' : : OQ : OQ' : : 1 : k. 

Suggestion. — Put x = p cos 6, y — p sin d, in the first equation ; it 
becomes 

p 2 - 2 p (a cos 6 + b sin 6) + a 2 + b 2 - r 2 = 0. 
Solving this quadratic in p, its roots are OP and OQ. Similarly with 
the second circle, etc. 

63. Find the circle through the origin and the intersections of 



+ j/ 2 + 3.r + 4?/ + 2 = 0, 



and 2z + 3t/+4 = 0. 

Also, the circle through (1, 2) and the same intersections. 

Ans. 2 (x 2 + y 2 ) + 4 x + 5 y = 0, 
2 {x 2 + y 2 ) - y -8 = 0. 



264 ANALYTICAL GEOMETRY 

64. Transform these equations to rectangular co-ordinates, pole at 
origin ; 

(1) p = a cos 0, 

(2) p = 2 a cos + 2 b sin 
in (1) multiply by p, 

.'. p 2 = a (p cos 0) 

.'. x 2 + y 2 = ax. 
in (2) p 2 = 2a (p cos 6) +26 (p sin 0) 

or, a: 2 + y 2 = 2 arc + 2 6?/, 

.*. (x - a) 2 + (y - 6) 2 - a 2 + b 2 . 

65. Two circles intersect in ; through O any line is drawn, cutting 
the circles in P and Q ; find locus of mid-point of PQ. 

Solution. — Take for the origin and let the circles be 
x 2 + y 2 + 2G x x + 2 ¥ t y = 0, 
x 2 + y 2 + 2G 2 x + 2 F 2 y = 0. 
Or in polar co-ordinates, 

p + 2 G t cos + 2 F l sin = 0,) Now let OP make 
p + 2 G 2 cos 6 + 2 F 2 sin 6 = 0. j /_B with x-axis. 
.-. OP = - 2 G v - 2 F x sin 0, 
and OQ = - 2 G 2 cos 6 - 2 F 2 sin 0. 

Let M be the mid-point of PQ and let OM = p. Then 2 p = OP + OQ. 
Hence the required locus is, 

2 /> = _ 2 (Gi + G 2 ) cos _ (2 F, + F 2 ) sin 0, 
or, /> 2 + (G 2 + G 2 ) (p cos 0) + (Fj + F 2 ) f/o sin 0) = 0, 

or, x 2 + y 2 + (G, + G 2 ) x + (P, + F 2 ) y = 0, 

i.e. another circle through O. 

66. Two segments AB and CD of a given line subtend equal angles 
at P ; find the locus of P. 

Suggestion. — Take the given line as a>axis and a point O on it as 
the origin. Let OA, OB, OC, OD be denoted by a, /3, 7, 5, respectively. 
Then the required locus is the circle, 

[a - - 7 + 5] {x 2 + y 2 ) + 2 (j3y - ad) x + yd {a- p) - ap (y - 5) = 0. 

67. Find the equation of a tangent to the circle x 2 4- y 2 = a 2 at the 
point (a cos a, a sin a). 

Solution. — Let (a cos /3, a sin fi) be another point on the circle. 
Then the chord joining the two points is, 



REVIEW EXAMPLES 265 

x y 1 

a cos a a sin a 1 
a cos /3 a sin /3 1 

or, x cos J (a + p) + y sin \ (a + /3) = a cos J (a - /S). 

Now put a = /3 and the chord becomes a tangent at the point (a cos a, 
a sin a) viz. : x cos a + ?/ sin a = a. 

68. Find the locus of the mid-point of chords of the circle x 2 + y 2 = a 2 , 
drawn through the fixed point (h, k). Take (a cos a, a sin a) and 
(a cos /3, a sin /3) as the extremities of one of these chords. Let (x, y) 
be the mid-point of the chord. 

Then we have 2 x = a (cos a + cos /3) 

2 y = a (sin a + sin /3) 
h cos i (a + /3) + A; sin ^ (a + /3) = a cos \ ( a— /3). 

Eliminating a, /3, from these three equations, we get for the required 
locus x 1 + y 2 = hx + A-?/. 

69. Given the base of a A ABC, and ab sin (C — a) = F, where a is a 
given angle. Find locus of the vertex. Take side AB as rc-axis, A the 
origin. Ans. The circle, x 2 + y 2 — ex — cy cot a + k 2 csc a = 
[where a, 6, c are the lengths of the sides]. 

70. A and B are fixed points ; M, N, are variable points on AB such 
that AM 2 + MN 2 + BN 2 = k ? . 

If PMN is an equilateral A, find the locus of P. 
Suggestion. — Take A as origin, AB as x-axis, let AB = a. 

Ans. The circle 2 (x 2 + y 2 ) -2 ax - ^ = fc a - a 2 - 

V3 

71. Find the equation of the chord joining the two points (2 a cos a, a) 
and (2 a cos /3, £) on the circle 

p = 2 a cos 0. 
^4ns. 2 a cos /? cos a = p cos (/3 + a — 0). 
Note. — The tangent to the circle, at the first point, is found by 
putting /3 = a in this result. Ans. 2 a cos 2 a = p cos (2 a — 6). 

72. Show that the polar of a point with regard to any circle of a co- 
axial system passes through a fixed point. 

Suggestion. — The polar of (h, k) with respect to the circle 
X 2 _J_ y i + 2 X x = c 2 
is hx + ky 4- X (x + h) = c 2 . 

The fixed point is determined by the equations 

x + h . = 0. Ax + &# = c 2 . 



266 ANALYTICAL GEOMETRY 

73. Find the circles through (1, 2), (1, 18) and touching the z-axis. 

Ans. x 2 + y 2 + 10 x - 20 y + 25 = 0, 
x 2 + y 2 - 14 x - 23 y + 49 = 0. 

74. Find the circle through (2, 0) which cuts the circles 

x 2 + y 2 = 4, x 2 + y 2 = 2y + 8 
at right angles. Ans. x 2 + y 2 — 4 z + 4 ?/ + 4 = 0. 

75. A, B, C, D, are four concyclic points and O is any other point. 
Prove OA 2 • area BCD + OC 2 • area ABD = 

OB 2 • area A CD + CD 2 • area ABC. 

76. The line Ax + By+ C = 0, cuts the circle 

x 2 + y 2 + 2G x x + 2 F,y + C, = 
in two points. Also the axis of x cuts the circle 

x 2 + y 2 + 2 G 2 x + 2 F 2 y + C 2 = 
in two points. Find the condition that these four points be concyclic. 

Suggestion. — The lines y = 0, Ax + By + C = 0, must meet on 
the radical axis. Ans. 2 C {G l - G 2 ) = A (C, - C 2 ). 

77. Find the co-ordinates of the limiting points of the circles 

x 2 + y 2 + 2 G v x + 2 Fjy + C x = 0, 
a: 2 + ?/ + 2 G 2 x + 2 F 2?/ + C 2 = 0. 

G, + X G 9 F, + X F 2 

_ Ans - x = — rrx"' y = --r+ir 

where X is either root of the equation 

(G x + G 2 X) 2 + (F t + F 2 X) 2 = (1 + X) (C, + C, X). 

78. Two chords of a circle cut at right angles show that the tangents 
at their extremities form a quadrilateral whose vertices are concyclic. 
Hence prove that the problem : To inscribe a quadrilateral in a circle 
whose sides shall touch another given circle is either indeterminate or 
impossible. 

79. A and B are points on the circle x 2 + y 2 = a 2 , and AB subtends 
a right angle at the fixed point (A, k). Find the locus of the mid-point 
of AB. 

Solution. — Let A be (a cos a, a sin a) and B (a cos /3, a sin /3). 

Then, the co-ordinates of the mid-point of AB are 

x = \ (a cos a + a cos jS) (1) 

y = i (a sin a + a sin /3) (2) 



REVIEW EXAMPLES 267 

Also the equation of a circle described on AB as a diameter is 
(x — a cos a) (x — a cos /3) + (y — a sin a) {y — a sin /3) = 0, 
and since this passes through the fixed point (A, &), we have 

(h — a cos a) (h — a cos /?) + (A: — a sin a) (Jc — a sin /3) = . (3) 
Hence we get the locus of the mid-point of AB by eliminating a and £ 
from equations (1), (2), and (3). The result is 

(x - h) 2 + (y - k) 2 + x 2 + 2/ 2 = a 2 . 
Show also that the locus of the foot of a ± from center of x 2 + y 2 = a 2 
on AB is the same circle. 

Also, find the locus of the intersection of tangents at A and B. 

Arts. The circle [h 2 + k 2 - a 2 ] (x 2 + y 2 ) - 2 a 2 hx - 2 a 2 ky + 2 a 4 = 0. 

80. Find the equation of the circle intersecting the circle 

x 2 + y 2 + 3 x + 5 y + 2 = 0, 
in the chord x + 2 y — 3 = 0, 

and the circle x 2 + y 2 — x + y — 2 = 

in the chord 2x + ?/ + 6 = 0. 

ilns. 3 (x 2 + ?/) + 5 ar + 7 ?/ + 18 = 0. 
Determine under what conditions such a problem is possible. 

Ans. The two lines must meet on the radical axis of the circles. 
Note. — This may be easily shown by Elementary Plane Geometry. 

81. Find the locus of a point such that the two pairs of tangents 
drawn therefrom to the circles 

x 2 + if + 2 Gx = 0, 
and x 2 + y 2 4- 2 Yy = 0, 

may form a harmonic pencil. 

Ans. The two parallel lines Gx — ¥y = -JL FG. 

82. The equations of two circles taking a center of similitude as 
origin may be written 

x 2 _|_ yi _ 2 ax + a 2 cos 2 ^ = 0, 
rr 2 + y 2 - 2 px + /S 2 cos 2 ^ = 0. 
Find the equation of the circle passing through the four points of 
contact of the common tangents from the origin. 

Ans. x 2 + y 2 - (a + /3) x + a@ cos 2 \p = 0. 

83. Is the point (1, 2) inside or outside the parabola y 1 = 8 x ? 

84. Find the points of intersection of 

y = 2 x — 4 a, 
and y 2 =4ax. Ans. (a, — 2 a), {4 a, 4 a). 



268 ANALYTICAL GEOMETRY 

85. Show that, y = x + 3 touches the parabola y 2 — 8 x — 8 = 0. 

Ans. Point of contact is (1, 4). 

86. Find the length of the chord made by the line 2 x + y = 8 on the 
parabola y 2 = 8 x. Ans. 6 V5. 

87. Find the co-ordinates of the focus, equation of directrix, and the 
length of latus rectum in the following parabolas : 

(1) y 2 = — 4 ax. Ans. ( — a, o), a: = a, 4 a. 

(2) a: 2 = — 4 a?/. .4ns. (o, — a), y = a, 4 a. 

(3) (y - l) 2 = 4 (re - 2). Ans. (3, 1), z = 1, 4. 

(4) x 2 + 4 y + 8 - 0. ^ns. (o, - 3), y + 1 = 0, 4. 

88. Find locus of intersection of tangents to parabola y 2 = 4 ax 
inclined at 60°. Ans. 3 x 2 - y 2 + 10 ax + 3 a 2 = 0. 

89. Find the chord joining the points of contact of the tangents, 

, a 

y = m,x -\ , 

y 1 Wj 

, a 
y = m 2 x H • 

^4ns. 2 (m x m 2 x -\- a) = y (m x + m 2 ). 

Suggestion. — The points are f — = , —J and [ — - , — ) . 

90. The line joining the point P on a parabola to the vertex cuts the 
I. from focus on tangent at P in Q. Find locus of Q. 

Ans. y 2 + 2 x 2 — 2 ax = 0. 

91. The equation of the tangents to the parabola from (h, k) may be 
written in either of the forms, 

h(y -k) 2 -k(y - k) (x - h) + a (x - h) 2 = 0, , 
or, (k 2 - 4 ah) (y 2 - 4 ax) = [ky - 2 a (x + h)] 2 . 

Suggestion. — See note on Joachimsthal's method in Appendix. 

92. If the intercept of the tangents from P on tangent at vertex is 
constant, find locus of P. Ans. An equal parabola. 

Suggestion. — If P is (h, k) intercept is \/k 2 — 4 ah. 

93. (x v 2/j), (x 2 , ?/ 2 ), (x 3 , t/ 3 ), and (x 4 , y 4 ) are the vertices of a re-entrant 
or concave quadrilateral whose sides touch the parabola. 

Prove x x x z = x 2 x 4 and y x + y 3 = y 2 + y 4 . 

94. Normals to a parabola are drawn at two points on opposite sides 
of the axes whose abscissae are in the ratio 1 : 4. Find the locus of their 
intersection. Ans. The curve 27 ay 2 = 4 (x — 2 a) 3 . 



REVIEW EXAMPLES 269 

95. Find the co-ordinates of the second point in which the normal at 
the point ( — p — J meets the parabola. 

a(2m 2 + l) 2 2a(2ra 2 + l) 
Ans. x = — 5 , y = 

Suggestion. — The normal is 

m 3 y + ra 2 (x — 2 a) — a = 0. 
Eliminate x from this and y 2 = ax. 

.'. m 2 y 2 + 4 amhj - 4 a 2 (2 m 2 + 1) = 0. 

— 4 a 2 (2 m 2 + 1) 
The product of the roots of this quadratic in y is, ■ —^ . 

But one root is — ; find the other by division ; etc. 
m 

96. Find the equations of the common tangents to the parabola 
if = 4 ax and the circle 2 (x 2 + y 2 ) — 9 ax = 0. 

Ans. 12 y -J- 16 £ -J- 9 a = 0. 

97. PQ is a normal chord of a parabola, F is the focus. Find the 
locus of the center of gravity of A FPQ. 

Ans. The curve 30 ay 2 (3 x - 5 a) - 81 if - 128 a 4 = 0. 

98. A circle cuts the parabola if =- 4 ax in four points whose ordi- 
natesare y lf y 2 , y.„ i/ 4 . 

Prove y t + y 2 + y 3 + y t = 0. 

Suggestion. — Eliminate x from f = 4 ax 
and (z - h) 2 + (y -k) 2 = r 2 . 

It gives a biquadratic in y which lacks the term in y 3 , etc. 

99. Find the locus of poles of chords of the parabola whose mid-points 
lie on the fixed line 

Ax + By + C - 0. 
Ans. The parabola A (if - 2 ax) + 2 a {By + C) = 0. 

100. (a„ /3,), (a 9 , /?.,), (a 3 , £,) are the feet of the normals from (h, k) 
to the parabola. Find a relation between them. 

Ans. (3 2 j8 3 (a, - o 8 ) + p. a /3, (o 8 - a,) + ^ /3 2 (a, - a 2 ) = 0. 

101. TP, TQ, are tangents to a parabola; TL is a _L to the axis and 
the _L from T on PQ meets the axis in M. Prove 2 LM = latus rectum. 

102. The major axis of an ellipse is divided into two parts equal to 
the focal distances of a point P on the ellipse. Prove that the distance 
of the point of division from either end of the minor axis is equal to 
the distance of P from the minor axis. 



270 ANALYTICAL GEOMETRY 

103. Find the co-ordinates of the mid-point of the chord which the 
x 2 y 1 
tf + b 2 



x 2 v 2 
ellipse —= + I2 = 1 intercepts on the line Ax + B?/ + C = 



A • C • a 2 B • C • 6 2 

Ans. 



a 2 A 2 + 6 2 B 2 a 2 A 2 + b 2 B 2 ' 

104. Determine C so that 

y = X + C 

may touch the ellipse 2 x 2 + 3 ?/ 2 = 1. 4 ns. C = -J- | \/l3. 

105. Find the tangents to the ellipse 

3 x 2 + 4 ?/ 2 = 12 
which cut off equal intercepts on the axes. 4ns. y ^j- £ = -J- \/7. 

106. OP, OQ are semi-diameters of an ellipse at right angles. 
Prove that PQ touches a fixed circle whose center is O. 

.Solution. — Let the ± from on PQ = p. Let a be the /_ which it 
makes with OP. Then 

cos a 1 sin a 1 

p = OP' p = OQ' 

square and add 

1 1 1 1 1_ 

p 2 ~ OP ? OQ 2 _ a 2 b 2 
hence p is constant, .*. etc. 

107. P is any point on the ellipse ; the _L bisectors of AP and AT 
meet AA' in M and M r . (A, A' are the vertices). 

Prove MM' = - (a 2 - b 2 ). 

Note. — This furnishes a method of describing the ellipse mechanically. 

108. PN and PM are the _Ls from a point P on two given oblique 
axes, /_ <p. 

If PN 2 + PAT = k 2 , find locus of P. 

Ans. Ellipse (x 2 + y 2 ) sin 2 <p = A- 2 , referred to its equi-conjugate 

diameters. 

The semi-axes a and b are determined by the equations, 

k 2 a 2 + b 2 b 

ctn 



2 a 2 

e 2 = 1 -ctn 2 ^ • 



REVIEW EXAMPLES 271 

109. Find the equation of a conic, having axes coincident with the 
co-ordinate axes and passing through the points (2 \/2, V3) and (4, 3). 
Find also its eccentricity. Ans. 3x 2 — 4 y 2 = 12 ; \ v /j 7. 

110. A straight line PQ subtends a right angle at each of two fixed 
points A and B. If P describes a straight line, show that Q describes a 
hyperbola whose asymptotes are _l_ to AB and the given line, respectively. 

111. P and Q are two points on an asymptote of a hyperbola, such 
that OP = 2 PQ [where O is center], and T is the point of contact of a 
tangent from P. The ^ PTP'Q being completed, show that its diagonals 
intersect on the curve. 

112. Two ellipses have the same focus and eccentricity and their 
major axes coincide in direction. PN and P'N" are the ordinates of the 
larger ellipse which are tangent to the smaller. Prove 

FP - FN = FP'-FN' (F is common focus). 

x 2 y 2 

113. The ellipse — + -f- = 1 slides between two rectangular axes. 

a 2 b 2 
Find the locus of its foci. 



Suggestion. — Let y = m.x + V m 2 a 2 + b 2 be one of the axes ; then 

put f or m to get the other. Now find the distances from a focus to 

m 

these axes [tangents]. Call these distances x and y respectively and 
eliminate m, etc. 

Ans. y 2 (x 2 -b 2 ) 2 + x 2 ( y 2 -b 2 ) 2 + 2 xy (x 2 -b 2 ) ( y 2 -b 2 ) = 4: xhf (a 2 -b 7 ). 

114. Find the equation of the normal to the conic xy = k 7 at the 

point ( A:X, -)i where is the angle between the axes of co-ordinates. 
Ans. (X 3 — X cos 0) x + (X" cos - X) y = k (X 4 - 1) 

115. If the normals at A, B, C, of the above conic meet on the curve, 
find the locus of the center of gravity of the A ABC. 

A ns. The conic 9 xy = k 2 cos ? <f> 



Part II 

SOLID GEOMETRY 



CHAPTER I 

THE POINT 

1. Fundamental ideas.* — A clear conception of the fol- 
lowing principles will facilitate an understanding of the 
Geometry of Three Dimensions. 

(1) In space, a straight line is regarded as the intersection 
of two planes. We may prove this statement thus : Since 
three points not in the same straight line determine a [one] 
plane, the intersection between two planes cannot contain 
three points not in the same straight line. Hence the inter- 
section is a straight line. 

(2) A point is the intersection of two straight lines, i.e., 
the intersection of a straight line with a plane which cuts the 
plane of the line, i.e., the intersection of three planes. Hence, 
a point in space will be determined by some relation to three 
given planes, a line by its relation to two planes, etc. 

(3) A plane curve, lying in space, is regarded as the inter- 
section between some surface or some solid with a plane. 
Thus, a circle might be determined [in space] by the intersec- 
tion of a plane with a sphere, or with a right circular cylinder, 
cutting the latter at right angles to its axis. 

2. Co-ordinates (rectangular). — The position of a point in 
space is determined by the intersection of three given planes. 
To assign convenient directions to these planes, three co- 



* The student will do well at this juncture to refresh his memory on 
the propositions concerning planes and straight lines in Solid Geometry. 

275 



276 



ANALYTICAL GEOMETRY 



ordinate planes are assumed, all perpendicular to each other 
[consider three faces of a cube forming one corner], and 
meeting in a point called the origin. The point is referred to 
these, and three planes are passed through it parallel, respec- 
tively, to the co-ordinate planes. The distances of these new 
» planes from the origin are the co-ordinates of the point. 




R 



r 



W 



M 



Fig. 151. 

Thus, let XOY, YOZ, and ZOX, be three planes cutting 
each other at right angles in the origin 0. These planes are 
called the xy, the yz, and the xz planes, respectively. Their 
lines of intersection, XX', YY', ZZ', are called the x, y, and z- 
axes, respectively. Let P be any point in space through which 
three planes have been passed parallel to XOY, YOZ, and 
XOZ, respectively. Then, by definition, its co-ordinates are 
evidently PQ, PR, PM ; or, more simply, ON, NM, and PM, 
the distances measured parallel to the x, y, and z-axes, re- 



THE POINT 



277 



spectively. The point P is now designated as the point 

0, y, z> 

Again, the co-ordinate planes divide all space into eight 
compartments called octants, which are numbered as follows : 
(1) - XYZ, (2) - YX'Z, (3) - X'Y'Z, (4) - Y'XZ. 
The last four go in the same direction around, below the xy 
plane. Let the student name them. The convention of signs 
follows the same rule as in Plane Geometry ; %. e., opposite 
signs for opposite directions. Thus, distances [co-ordinates] 
measured in the direction of OX, OY, and OZ are positive, 
while those in the direction of OX', OY', OZ' are negative. 
If (x, y, z) are the lengths of the co-ordinates of any point, 
they have the following signs in the various octants : 





I 


II 


III 


IV 


V 


VI 


VII 


VIII 


X 


+ 


— 


— 


+ 


+ 


— 


— 


+ 


y 


+ 


+ 


— 


— 


+ 


-f- 


— 


— 


z 


+ 


+ 


+ 


+ 


— 


— 


— 


— 



These should be verified by the student. 

3. To locate a point P. — In Pig. 151, § 2, measure 

OX = x [on x-axis]. 

NM = y [parallel to ?/-axis]. 

MP = z [parallel to z-axis]. 
The lengths ON, OS, OT may also be taken as the co- 
ordinates of P. 

Exercise. — In which planes do the following points lie : (1) (x, y, o), 
(2) (o, o, o), (3) (x, o, o), (4) (o 5 y, z), (5) (x, o, *), (6) (o, o, *), (7) (o, y, o), 
etc.? 



278 



ANALYTICAL GEOMETRY 



4. Polar co-ordinates. — The point P may also be deter- 
mined in space as follows : 

Draw OP, and through it pass a plane _L to the xy plane 

and cutting the latter in 

the line OM. The position 

of P is then known if the 

line OP, the radius vector, 

and the angles <f> and 0, the 

vectorial angles, are given. 

These three magnitudes 

together constitute the polar 

co-ordinates of the point P. 

For transformation from 

rectangular to polar co-or- 

/ dinates and vice versa, we 

* 152 ' have, 

x = ON = OM cos <£ = OP sin 6 cos <f> = p sin 6 cos <f>. 

Similarly, we get y and z. 

x = p sin cos (f> , "I 
y = p sin 6 sin <j> , i- 
z = p cos 0. J 

Also, 




p = y/tf+Y 



tan d> = — , 
x 



+ * 2 ,1 



tan — 



vx° 



5. Direction angles. Relation among them. — The angles 
which the radius vector of a point P makes with the axes are 
called its direction angles ; and their cosines, the direction 
cosines. They are generally denoted by the Greek letters, 
a, ft y. 




279 



Fig. 153. 

By projecting OP on the three axes, we get, 

x = p cos a , ^ 
y = P cos ft L 

Z = p COS y. J 

.% x 2 + y 2 + z 2 = p 2 (cos 2 a + COS 2 (3 + COS 2 y). 

.'. COS 2 a + COS 2 /? + COS 2 y = 1. 

Note. — The co-ordinates of a point are its projections on the axes. 

Again, the direction cosines of the radius vector of any 
point (xj y, z) are obviously, 

x x \ 

cos a = 



p vx 2 -J- y 2 H- z 



cos B = £ = 



COS y 



Hence, (1) If any three real numbers are divided by the 
square root of the sum of their squares, the results are the 
direction cosines of some line, or of the radius vector of some 
point. 



i80 



ANALYTICAL GEOMETRY 



(2) The direction cosines of the radius vector of any point 
are proportional to the rectangular co-ordinates of the point. 

Example 1. — Find the direction cosines of the radius vector of the 
point (1, 2, 3). An ^ _J_ ^2_ _3_ 

vii' vii' vii' 

Ans. 



Ex. 2. —The point (1, 0, 3). 



-L, o,-L 

V10 V10 



Hence, its radius vector is perpendicular to the y-axis, a 
fact which is also seen by observing that the point lies in the 
xz plane. 

Ex 3. The direction cosines of a point are proportional to 3, 4, and 5. 
What are they ? 3 4 5. 

nS ' V50 V50 V50 
Observe that these results may be simplified. 

Note. — The projection of a given line on another line is equal to the 
product of the former by the cosine of their included angle. 

6. The projection of the join of the ends of a broken 
line on a given line is equal to the sum of the projections 

B 




M N 

Fig. 154. 

of the parts of that line. — Let the broken line be ACDB, 
and XX' the line on which it is to be projected. 

Let 6 V 6 2 , <9 3 be the angles made by AC, CD, DB with XX'. 

Then LP = LM + MN 4- NP, , 

or, AB cos 6 = AC cos 6 X + CD cos 2 + DB cos 3 . 

.-. , etc. 



THE POINT 



281 




Note. — If any portion of the broken line has a negative direction, its 
angle is also negative, and therefore the product of its length by the 
cosine is always positive. This important principle of projections de- 
pends on the convention of signs, or the fact that the algebraic length of 
a line and the cosine of its angle always have the same sign. 

7. Angle between two lines in terms of their direction 
cosines. — Let OP 1? OP 2 , be two radii vectores parallel respec- 
tively to two given lines in space, whose direction angles are 
respectively (a v fi v 7l ), (a 2 , (3 2 , y 2 ). 

Z P 




Fig. 156. 

Then, the projection of OP t on OP 2 
broken line ONMR on OP,. 



projection of the 



ft cos = ON cos a 2 -f- MN cos fi 2 + MP X cos y 2 
= (p L cos a x ) cos a 2 + (p t cos /?.) cos j3 2 
+ (ft COS yj COS y 2 



2S2 ANALYTICAL GEOMETRY 

[since the co-ordinates of P x are the projections of p x on the 
axes]. 

.'. COS 6 = COS a t COS a 2 + COS ft COS ft + COS y t COS y 2 , 

Which gives the cosine of the angle between the two lines 
in terms of their direction cosines. 

Discussion. — (1) If the given lines are parallel, 

a i = *21 ft = &> 7i = 7 2 - 

(2) If the lines are _L, we have 

cos aj cos a 2 + cos ft cos ft + cos 7 X cos y 2 = 0. 

EXERCISES. 

1. (x v y v zj, (x 2 , y 2 , z 2 ), are two points in space, d the distance be- 
tween them, and a, ft 7 the direction angles of the line joining them. 

Prove, d = V(x 2 - x,f + (y 2 - y x f + (z 2 - z,)\ 

Also, d cos a = x 2 — x v etc. 



.f 2 


— 


x \ 




d 




y* 


- 


Vi 




d 




z 2 


- 


z i 



cos/3 

cost- d 

2. The square of any line is equal to the sum of the squares of its 
projections on the axes. 

3. If (x, y, z) divides the distance between the points in Ex. 1 in the 
ratio of ra : n, prove, 



y = 



n 


+ m 


™Ji 


+ my 2 


n 


+ ra 


nz l 


+ mz 2 



n + m 



4. If (#, y, z) is the mid-point, prove, 

2 = 2 ( X 2 + X l)l 

y = h(y-2 + 2/i)» 
z - § (z 2 + z x ). 



THE POINT 



283 



5. Find the distance between the points (2, 3, 6), (1, 2, 4). Between 
(1, 3, 5), (1, 2, 3). 

6. Show that the triangle whose vertices are (1, 2, 3), (2, 3, 1), 
(3, 1, 2), is equilateral. 

7. The projections of a line on the axes are 2, 6, 6. What is the 
length of the line ? See Ex. 2. 

8. If the origin is moved to the point (h, k, I), the axes remaining 
parallel to the old ones show that the formulas for transformation are 

x = x' + h, y = y' + k, z = z' + I. 

9. If the origin remains unchanged, and the axes are turned so that 
they make these direction angles with the old axes (a x , § x , 7J, (a 2 , j8 2 , y 2 ), 
(a 3 , jSj, 7 3 ), prove that the formulas for transformation are as follows : 

x = x' cos a l + y / cos a 2 + z' cos a 3 . 
y = x ' cos |8 1 + y' cos j3, + z' cos j8 3 . 
z — x' cos 7j + if cos 7, + z' cos 7 3 . 




x = ON, 7/ - MN, z = PM. 
x' = ON', y' = M'N', 2' = PM'. 

Now, the projections of OP and ON'M'P on the z-axis [old] are equal 
Similarly on the other axes. Hence, the above formulae. 

10. Find the radius vector with its direction cosines for each of the 
points (2, 4, 6), (3, -2, -1), (1, -2, -3). 



284 ANALYTICAL GEOMETRY 

11. The direction cosines of a line are proportional to 2, 6, and 7. 
What are they ? 

12. Two direction angles of a line are [60° and 45°] ; [60° and 30°] ; 
[135° and 60°]. Find the third angle in each case. 

Ans. (1) [60° or 120'] ; (2) [90°] ; (3) [60° or 120°]. 

13. Find the angle between the lines whose direction cosines are 
proportional to (2, 3, 5) and (— 3, 2, —1). 

14. Find the rectangular co-ordinates of the points 



a) 



( 2 > v 1} v (*. J. J). 



[Ans. (2) 1, V3, 2\/3]. 

15. Show that the distance between two points in terms of their 
polar co-ordinates is equal to 

d = v/o x 2 + p 2 2 — 2 p x p 2 [cos (0 X — 6 2 ) sin <p t sin <£ 2 + cos <f> y cos 2 ]. 

16. Find the co-ordinates of the point which divides in the ratio of 

1 : 3 the distance between (1, - 2, 3), (2, 3, 5). 

17. Find the co-ordinates of the point which divides in the ratio of 

2 : 5, the distance between the points (1, 4, 6,), (— 2, — 3, —6 ). ' 

18. A line makes equal angles with the co-ordinate axes. Show, 

cos a = cos j8 = cos y = — — • 
V3 

19. Find the polar co-ordinates of the point (2, 3, 6). 

20. Prove by co-ordinates that the lines joining the mid-points of the 
opposite sides of any quadrilateral pass through a common point and are 
bisected by that point. 



CHAPTER II 
THE PLANE 

8. Normal equation of the plane. — Let OX be the_L from 
the origin to the plane. Let P be any point in the plane and 
OP its radius vector. Also, OX =_p. The direction angles 




Fig. 158. 

of OP are a, fa, y, and its co-ordinates [OR, RQ, QP], x, y, and z. 
Then, the projection of OP on OX = projection of ORQP on 
OX. But OX is the projection of OP on OX since OX is 
_L to plane. 

.'. x cos a -\- y cos /3 -f- z cos y = p 

is the required equation of the plane. 

Discussion". — (1) If the given plane is perpendicular to one of the 
co-ordinate planes, e.g., the yz plane, then OX lies in the yz plane, and 
a = 90°, cos a = 0, and the equation becomes, y cos /8 + z cos y = p. 

(2) If parallel to one of the co-ordinate planes, xy, for example, ON 
lies in the z-axis, hence 

cos a = 0, cos /3 = 0, cos y = 1. 
.-. z = p. [Equation of the plane.] 
285 



286 ANALYTICAL GEOMETRY 

Note. — The reader will now see that in locating a given point, e.g., 
(a, 6, c), we are really finding the intersection between the three planes 
[x = a, y = 6, z - c]. 

EXERCISES. 

Describe each of these planes : 

(1) x cos a + y cos /3 = p, (3) x = p, 

(2) x cos a + z cos 7 = p, (4) 3/ = p. 

9. Every equation of the first degree in three variables 
represents a plane. — Let the general equation of the first 
degree be 

Ax + By + Cz = l> (l) 

Now, the expressions 

ABC 



VA 2 + B 2 4- C 2 VA 2 + B 2 4- C 2 VA 2 + B 2 4- C 2 

are the direction cosines of some line. Hence, dividing (1) 
by VA 2 + B 2 + C 2 , we obtain, 

Ax B?/ Qz 

+ f ..-_ 1. = + 



VA 2 +B 2 -}-C 2 VA 2 + B 2 + C 2 VA 2 + B 2 -f- C 2 

- , D (2) 

VA 2 4. B 2 + C 2 

which is the normal form of the equation of a plane. Hence, 
every equation of the first degree can be reduced to the nor- 
mal form, and therefore represents a plane. 

Discussion. — (1) p = — = JL from origin to (1) [plane]. 

VA 2 +B 2 +C 2 

(2) To construct plane (2), draw the radius vector of the point 
(A, B, C), and a plane ± to it at that point is the plane required. 

(3) To reduce (1) to the normal form, make D positive and divide by 
VA 2 + B 2 + C 2 . 

(4) A plane parallel to Ax + By + Cz = D is Ax + By + Cz = h 



THE PLANE 287 



EXERCISES. 

Interpret the following equations : 



1. 


3x + 4y = 


. 6. 


2. 


5x + 2z = 


9. 


3. 


2y- hz = 


10. 


4. 


x + y = 0. 




5. 


2x - Sz = 


0. 


6. 


iy + 2z = 


16. 


7. 


x = ±3. 




8. 


z = 0. 




9. 


2/ = ±6. 




10. 


7/ = 0. 




11. 


2 = i 5. 





Ans. A plane JL to xy plane. 
Ans. A plane ± to 2:2 plane. 
Ans. A plane _L to yz plane. 
12. 2 = 0. 



13. 


xyz = 0. 


14. 


xy = 0. 


15. 


xz = 0. 


16. 


yz = 0. 


17. 


2 ?/ - 2 = 0. 


18. 


Z + 2 = 3. 



10. Symmetrical equation of the plane. — This is the 
equation in terms of a, b, c, the intercepts of the plane on the 
axes. 

Let the required equation be, 

Ax + B?/ + Cz = D. 
Then, since this plane passes through the points (a, o, o,), 
(o, #, o), (o, o, c), we have, 





Aa 


= D, 


A = 


D 

: — j 

a 




Bb 


= T>, 


B = 


D 




Cc 


= D, 


C = 


D 


Whence, 


x y 


z 

+ -=: 


1 


is the 


required equation 


l of the 


plane. 





U. Angle between the two planes A 1 ic- r -B 1 2/+C 1 «+D 1 =0, 

and A 2 x -+- B 2 y + C 2 z + D 2 = 0. — The angle between two 
planes is evidently equal to the angle between their normals 
from the origin. 



288 ANALYTICAL GEOMETRY 



The direction cosines of the latter are 

A t B, C, 



VA ; 2 + B, s + C, 2 V V . . . 

A2 -t> 2 ^2 



VA 2 2 + B 2 2 + C 2 2 V ' V 

Hence, the angle 6 between the two normals and .'. between 
the planes is determined by 

A t A 2 + B X B 2 + C t C 2 



cos 



VA, a +- B x 2 + C x 2 • VA 2 2 + B 2 2 + C 2 2 

Discussion. — (1) If the planes are parallel, their normals have the 
same direction cosines ; and since the latter depend only on the coeffi- 
cients of x, y, and 2, these coefficients in both equations must be equal or 
proportional. Hence, the condition for parallelism is, 
A, _B 1 _C 1 

This result is also obtainable by putting 8 = 0, cos = 1, above. 
(2) If the planes be perpendicular, cos 8 = 0, 
.-. A,A 2 + BjB 2 + CjC, = 0. 

12. Distance from a point to a plane. — Let the given 
point be (x v j v z x ), and the plane, 

x cos a + y cos /? + z cos y = p. 
Then the plane x 1 cos a + y 1 cos ft -f- z x cos y = p x 
passes through the given point and is parallel to the given 
plane. 

But d = p x — p = x x cos a + y x cos ft -\- z t cos y — p. 
.-. d = x x cos a + y 1 cos /3 -f 2j cos y — p. 

If the given plane is Ax + By -f- Gz = D, 

we have a = — , 

VA 2 + B 2 + C 2 

Note. — If this formula give a positive result, the given point and the 
origin are on opposite sides of the given plane ; if negative, on the same 
side ; if the numerical distance only is desired, the sign is immaterial. 



THE PLANE 289 

EXERCISES. 

1. Interpret these equations : 

(1) y = mx + c. (4) 2 x + 3 y = 0. 

(2) z = nx + c. (5) a; - z = 4. 

(3) t/ = m2 + c. (6) y + 3z = 1. 

2. Reduce to the normal and symmetrical forms, and find which 
octant each of these planes cuts : 

(1) 4 x - 5 y + 2 z = 9. (5) - a - 2 y - 2 = 10. 

(2) 3x + 4y - z =5. (6) - a; + y - z = 4. 

(3) 5 x + 5 y + z = 2. (7) - 2 z + 3 y + z = 8. 

(4) 2y-3:r+ z =4. (8) 2 y - x - z = 3. 

3. Find the intercepts of the following planes : 

(1) y + 2 x - z = 3. (4) - x + y - z= 2. 

(2) 2x- 5y + 3z = 5. (5) a + 2 y - 5 z = 14. 

(3) a; + y + z = 7. (6) 2 z - 3 y - z = 10. 

4. A plane is 5 units from the origin and ± to the line whose direc- 
tion cosines are proportional to 2, 3, 5. Find its equation. 

5. Find the distance from the origin to the plane whose intercepts 
are [1, 5, 9]. 

6. Find the equation of the plane passing through the points 

(0, - 2, 4), (2, 2, 2), and (1, 0, 3). 

7. Find the angle between the planes 
x — 2y + 4 z = 6j 



3z-y-2 = 0' (1) 



and that between the planes 

2x-y + 3z = l 
x — y — 2z = 3 



(2) 



8. Show that the angles which the plane Ax + By + Cz = ~D makes 
with the co-ordinate planes are 

C 

(1) cos-i 

v VA 2 + B 2 + C 2 

B 

(2) cos-i 



(3) cos-i — = 



290 ANALYTICAL GEOMETRY 

9. Find the distance from the point (1, — 2, — 3) to the plane 
2x -y + 32 = 4. 

10. Show that the plane A ( x — xi) + B (y — yj + C {z - z t ) = 
passes through the point (x x , y 1 , z ± ) and is parallel to the plane 
Ax + By + Cz = T>. 

11. Find the equation of the plane through the point (2, — 3, 5) and 
parallel to the plane x + 2y — Sz = 2. 

12. What three equations must be satisfied in order that the plane 
Ax + By + Cz = D may pass through the points (x x , y v Zj), (x 2 , y 2 , z 2 ), 
and (x 3 , y 3 , z 3 ) ? If the equations differ only by constant factors, what 
about the three given points? 

13. Find the equation of the plane through the points (2, 0, — 3), 
(1,0, — 2), and ± to the plane 

2x-y+2z = 4. 

14. Find the plane through the points (1, 2, 0), (0, 3, - 2), 
(-2,0, -4). 

15. Find the plane through the point (2, 2,-1) and ± to each of the 
planes, 

y — x — 3z = 3, 
2y + x+ z = 9. 



CHAPTER III 

THE STRAIGHT LINE 

13. A straight line in space is generally represented by 
two equations of the first degree considered as simultaneous, 
i.e., by the intersection between two planes. The equations 
of any two planes through a straight line are sufficient to 
determine the line. The simplest planes, however, available 
for this purpose, are two planes [through the line] which are 
perpendicular to two of the co-ordinate planes, and called the 
projecting planes of the line. Thus, 

x = mz + b, 
y — nz + c, 

are two planes representing a straight line ; and since they 
are _L to the xz and yz planes respectively, they are two of its 
projecting planes. 

14. Equations of the straight line through a given point 
and in a given direction. — Let (x v y lt z x ) be the given point 
and a, /?, y, the given direction angles of the line. Also, 
(x, y, z) is any point on the line, and d the distance between 
the two points, 

Then, d cos a = x — x x , 

d cos fi = y - y 13 

d COS y = z — Z x . 



x-x x _y-y 1 _z -z y 



COS a COS /? COS 

are the equations of the line. 

291 



(i) 



292 



ANALYTICAL GEOMETRY 



15. Equations of the straight line through two given 
points (x v y v zj and (x 2 , y 2 , z 2 ). — If the line (1), § 14, passes 
through the point (x 2 , y 2 , 2 2 ), we have, 



'i 



2/2 



f3 " COS y 



COS a COS p 

Dividing (1) by (2), member by member, we obtain, 



x -x x y -y x 



2/2-2/i 



z — z x 

Zn Z-, 



(2) 



(3) 



as the required equations of the straight line through the two 
points. 

16. To find the projecting planes of a given line. — Let 

the line be given by the equations, 



A lX + B t y + O lZ + D x = 
A^ + B 2 y + C 2 z + D 2 = 



or, for brevity, 

Then, the equation, 
represents any plane passing through the inter 



mB 1 4- nS 5 



(i) 

(2) 
(1) 

(2) 

(3) 

-section of the 

planes (1) and (2). (Why ?) Now, if m and n be so chosen 
that one of the variables will vanish, then (3) will represent 
one of the projecting planes of the given line. 

For example, if z vanish, the resulting equation (3) will 
represent the plane through the line which is J_ to the xy 
plane. Similarly for the other projecting planes. In prac- 
tice, however, we simply eliminate from the two given equa- 
tions the necessary variable. 

17. To find the piercing points of a given line on the co- 
ordinate planes. — Let the given line be represented by the 
equations, 



THE STRAIGHT LINE 293 

x = mz -f- b n\ 

y = nz + c (2) 

To obtain its piercing point on the xy plane, we make z = 
in (1) and (2), and solve the resulting equations for x and y 
The result is x = b, y = c. Hence, the line meets the xy 
plane in the point (b, c, o). Similarly for the remaining 
piercing points. 

18. Condition that two lines may meet Let the lines 

be given thus, 

x = mz + b ) 

y = nz + c) W 

x = m x z -f b t \ 

y = n x z-\-c x \ ( 2 ) 

Suppose they meet. Then, to find their point of intersection 
we must solve these equations for x, y, z. But there are four 
equations and only three unknown quantities. Hence, the 
condition that the lines meet is that the co-ordinates satisfy- 
ing any three of the equations should also satisfy the fourth. 
Equating the values of x , and also those of y, then solving 
each equation for z, we obtain, 

z _ \-b 

m — m l v9 

c. — c 

z = ^^, w 

Hence, if the lines intersect, we have, 

b x — b c 1 — c 

m - m 1 ~ n-n, • • • • • ( 5 ) 
which is the required condition. 
Note. — A line in the form 

x -*i y - y, g--gi 
A B C~ 



294 ANALYTICAL GEOMETRY 

can be reduced to the typical form of § 14 by dividing the denominators 
by VA 2 + B 2 + C 2 , for the results are the direction cosines of the line. 

19. Angle between two lines. — Let the given lines be ; 

x - z t _ y — y x _ 3. - Zi „, 

(2) 

(1) 
(2) 

W+Bi'+Cj 2 v . . . . v r ... 

Hence, the angle between them is determined by, 

AA, + BB. + CC t 

VA 2 + B 2 + C 2 . VA t 2 + B x 2 + C x 2 ' 



ABC 






x-x 2 y -y 2 z-z 2 






K B, t 






Then the direction cosines of the lines are, 






A B 


c 




Va 2 + b 2 + c 2 ' V . . . . V 


. 


. 


A, B t 

J ■ - - ) r- 


o. 





COS V = 



20. Angle between a line and a plane. — Let these be 

given as follows : 

s - *i _ y -Vx _ g - 5 m 

A ~ B C w 

A x x + B l2 / + C t 2 = D (2) 

Now, the angle between the line and the plane is the com- 
plement of the angle between the line and the normal to the 
plane. 

The normal from the origin to the plane is 









X 


y 

~B. 


z 




. • • • 


. . (3) 
[Why?] 


If <£> is 


the 


ang 


;le between (1) 


and 


(2), 


and the 


angle be- 


tween (1) 


and 


(3), 


sin <f> = 


COS Br 











THE STRAIGHT LINE 295 

AA. + BB. + CC 1 
.*. sin <b = _ , 

VA 2 + B 2 + <J 2 • VA^ + B^ + C^ 

Discussion. — (1) If the line be parallel to the plane sin = 0, 
.-. AA, + BBj + CC X = 0. 

(2) If ±, sin = 1, whence, 

A-.?. - c 

A^~ B^ ~ C^" 

(3) If the line lies in the plane, the angle between them is 0, and the 
plane passes through the point (x l , y lt z,). 

.-. AA t + BB, + CC t = 0, 
and Aj x t + Bj y t + C, z x = D. 

Note 1. — Referring to the result of § 19, if the lines are parallel, 

A _5. -SL 

A, ~ B, ~ C : ' 
and if JL, AA 1 + BB, + CC, = 0. 

Note 2. — The _L from the point {x x , y 1 , z x ) to the plane 
A x x + Bj y + C t z = D is evidently 
x -x, y - ?y, g_~-_g, 
A, B, C, ' 

for this line passes through the given point, and has the same direction 
cosines as the normal from the origin to the given plane. 

21. If a line in space is ± to a plane, its projections are 
perpendicular to the traces* of the plane. — Let any plane 
be Ax + By + Cz = D. 

Then its traces are, 

Ax + By = D (1) 

By + Cz = B . . . . . . . (2) 

Ax + Cz = D (3) 

Also, the _L from the origin to the plane is 
x _ ?/ _ z 
A~B~ C' 



* The traces of a plane are its lines of intersection with the co-ordi- 
nafe planes. The projections of a line are the lines of intersection of its 
projecting planes with the co-ordinate planes. 



296 ANALYTICAL GEOMETRY 

and its projections are 

Oy - Bz = j ...... . (4) 

Cx -Az = 0j 
But the lines (4) are evidently _L [respectively] to (1), (2), 
and (3). .*., etc. 

22. To pass a plane through a given point and a given 
line. — Let these be given thus : 

(x 2 , ij 2 , z 2 ) and A = B = -g— * • 

Now let the required plane be 

A;c + By + Cz = D (1) 

Then, since it must pass through the given point, 

/. Ax 2 + By, + Cz z = D (2) 

and through the given line, 

.-. AA^BBj + CC^ (3) 

and Ax t + By x + Cz x = D (4) 

Now eliminate A, B, C, and D from (1), (2), (3) and (4), 
etc., to obtain the equation of the required plane. 

EXERCISES ON CHAPTER III. 

1. Find the positions of the following lines : 
3x+4?/-z=4) 2x - y = 1) y = 2) x = 1 ) x = -1) 



x -2y + Sz = 10) Sy -2z = 3j z = -3)' y =4) z = 3}' 

2. Find the equation of the line through the points (2, 4, — 5), 
(3, — 2, — 9). Determine its piercing points. 

3. Two projecting planes of a line are 

2z-3y -3 = 0,) , 

* . 'J Find the third, 
a:— 2s — 4 = 0. 



THE STRAIGHT LINE 297 

4. Find the angles between the lines 

x _ y _ z 

2 ~~ 3 ~ 
, x y z 

and 5-^-r 

Also, between the lines 

2z + y -2x -S = 0, ) 

3?/ + z -3x - 5 = 0,) 

and ?/ — 3 x — 4 = 

22 - 5a; - 5 = 

5. Show that the lines 

3x + 2y + z-» = 0.) d 2x- y- 2 = 0,) 
x + 2/ -2z - 3 = 0,j ' 7z + 10?/ -82 = 0,J 
are perpendicular. 

6. Find the equations of the line through the point (1, — 2, — 5) and 
|| to the line 

2 - y + 2 x + 3 = 0, 
z - 3 x + 5 ?/ - 1 = 0. 

7. Find the equations of the line through the point (1, 2, 3) and ± to 
the plane 

x + 4y -2« -8 = 0. 

8. Prove that if two lines are parallel, their projections are parallel 

9. Find the angle between the line 

x - 1 _ y + 3 _ 2-6 
2 - ^T~ -3 
and the plane 3a; - 4y + bz — 2 = 0. 

10. Show that the three planes, 

2x -By +2 + 1 = 0,1 
5.r + 2- 1 = 0, L 
19 x - 3 y - 4 2 - 5 = 0, J 
are concurrent in one straight line. 



CHAPTER IV 

SURFACES. SURFACES OF REVOLUTION 

23. A single equation in three variables, some of which 
may be absent, represents a surface. — Take the equation 
/ (x f y, z) = 0. This expresses the condition satisfied by the 
co-ordinates of all points on its locus, i. e., the law governing 
the motion of a point in space. Now these points [or the 
various positions of one moving point], as it were, cannot lie 
scattered indiscriminately in space, since at any moment their 
co-ordinates must satisfy the given equation. Hence, the 
given equation cannot represent a solid. It therefore repre- 
sents a surface. 

Again, take / (x, y) = 0. In the plane xy this equation 
represents a curve. Now, if a line _L to this plane move 
parallel to itself, so that one of its extremities traces out the 
curve in the plane, the line will generate a cylindrical surface, 
and it is obvious that the co-ordinates of any point on the 
moving line always satisfy the given equation. Hence, a 
single equation in two variables represents a cylinder. 

Finally, take f (x) = 0. This evidently represents a plane 
J_ to the z-axis, or, when its sinister is factorable, it repre- 
sents a series of parallel planes. .*. , etc. 

EXERCISES. 

What loci in space are represented by the following equations ? 

(1) x 2 + y 2 =25. (3) y 2 = 12 z. 

(2) z 2 + x 2 = 9. (4) (x - 2) 2 = 12 (y + 3) 2 . 



SURFACES OF REVOLUTION 



299 



Wir + r- 1 - 



(6) 



25 



1. 



(7) 


x 2 - 


-y 2 = 


= 0. 




(8) 


re 3 - 


->2x 2 


- 5rc+ 6 = 


= 


(9) 


(x- 


- 4) (y + 3) - 0. 




(10) 


xyz 


(x 2 - 


2z+ 1) = 


0. 



24. Surfaces of revolutions. Definitions. — These sur- 
faces are generated by the revolution of a curve, the generatrix 
about a fixed axis, the axis of revolution. A section of the 
surface by a plane through this axis is called a meridian 
section. It is clearly equal to the generating curve. A sec- 
tion _L to the axis is a right section, and by definition is a 
circle. The traces of a surface are its intersections with the 
co-ordinate planes. 

25. General equation of a surface of revolution. — Let 

ABC be the generating curve [in the plane xz], and let the 




z-axis be the axis of revolution. Let BDE be any right 
section of the surface. Put CB = a [= CF]. 

Then, for any point on this section, x 2 + if = a 2 . 

But a is the x of the point B before the rotation. 



300 ANALYTICAL GEOMETRY 

Hence we must find x from the equation of the generating 
curve / (x, z) = 0, and substitute for a. 

.:x* + tf=f(z) ... (X) 
is the equation required. 

If the revolution is about the ?/-axis, we have 

x 2 +z 2 =f(y), etc. 
We shall use equation (A) as the general equation of a 
surface of revolution. 

26. Equations of various surfaces. — 

(1) The sphere. 

Here, generatrix is x 2 4- z 2 = r 2 . 

r.x 2 = r 2 - z 2 =/(«). 
.'.x 2 + y 2 +z 2 = r 2 
is the equation of the sphere.* 

(2) The paraboloid of revolution. 
Here, generatrix is x 2 = 4 az. 

.'. f (z) = 4 az, . '. x 2 -f- y 2 = 4 az 
is the equation of the paraboloid of revolution. 

(3) The ellipsoid of revolution. 

(a) Prolate spheroid. 

x 2 z 2 
Here, generatrix is— -\- — =1 [revolution about major axis]. 

• ?1 _u y l _i_ t - 1 

* ' b 2 " "*" V- "*" a 2 ~ 

is the equation required. 
(/?) Oblate spheroid. 

Here, generatrix is — + — =1 [revolution about minor axis]. 
a i b l 



* If center is at (h, k, I), the equation becomes 

(x - hf + (y - k) 2 + (z - If = r\ 



SURFACES OF REVOLUTION 301 

x 2 y 2 z 2 

.'. f- — H — =1 

a 2 a 2 b 2 

is the equation required. 

(4) The hyperboloid of revolution. 

(a) Hyperboloid of one sheet [revolution about conjugate 

axis]. Its equation is found to be 

x 2 y 2 z 2 _ 
-£ -r 2 — Th — l, 
a 2 a b 2 

either directly, or by putting — b 2 for b 2 in the equation of the 

oblate spheroid. 

(/?) Hyperboloid of two sheets [revolution about transverse 
axis]. 

Its equation is 

x 2 y2 z 2 



b 2 b 2 a 2 
(5) The cone of revolution. 
Here, generatrix is z = mx + c. 



■••/(«) = 



m 



.-. m 2 (x 2 + if) = (z- cf 
is the equation required. 

27. Discussions of these equations. — 

(1) All plane sections || to the co-ordinate planes are circles. 
Also, the equation remains unaltered by any rotation of the 
axes. .*. all plane sections of a sphere are circles. 

(2) All plane sections || to the xz and yz planes are par- 
abolas. Those || to the xy plane are circles. 

(3) (a) The surface lies between the tangent planes 

x = ± b, y = ±b, z = ± a. 
Its traces on the xz and yz planes are ellipses. 



302 



ANALYTICAL GEOMETRY 



(/?) Surface lies between the tangent planes 

x = ±a, y = ia, z = ±b. 

Traces on the xz and yz planes are ellipses. 

jf ^ _ a> the equation reduces to that of the sphere. 

(4) (a) Sections || to the yz and xz planes are hyperbolas. 
The smallest circular section [|| to xy plane] is the trace on 
the xy plane, and is called the " gorge circle." 

The planes x = ± a, each cut two intersecting straight lines 
on the surface. 

(/?) Sections || to xz and yz planes are hyperbolas. 

(5) Sections || to the xz and yz planes are hyperbolas. 
If c = 0, the equation becomes, 

m (z 2 + y 2 ) = 2\ 

Hence, any section parallel to the axis of the cone is an 
hyperbola ; any section through the axis consists of two inter- 
secting straight lines. 



Sections of a cone. 




We propose to determine the 
nature of any section of a cone 
by finding its equation referred 
to a pair of axes in the secant 
plane. 

APB is any section of a cone, 
through the y-Sixis and .-. _L to 
the xz plane. 

Let OB and OY be the axes 
of reference. Also P is any 
point on the section, whose co- 
ordinates in space are (x, y, z) ; 
and referred to OB, OY, in 
plane ABP, its co-ordinates 
are (x 19 y x ). 



SURFACES OF REVOLUTION 303 

Let Z XOB = <£, Z OKZ = 0. 

Then, draw PM _L OB, MN J_ OX. 

Now, ON = OM cos cj>, MN = OM sin <£, 

or x — x x cos <£, 2; = x t sin <£, 

2/ = FM = y [since PM JL #2 plane]. 
Substituting these values for x, y, z, in the equation of the 
cone, [m 2 (x 2 + ?/ 2 ) = (z - c) 2 ] ; 

and noting that m = tan 6, we obtain 

tan 2 [x 2 cos 2 <£ + t/^] = [x t sin <£ - c] 2 . 

Or, performing indicated operations, and omitting the sub- 
scripts, we obtain, 
y 2 tan 2 $ + z 2 [cos 2 <£ tan 2 - sin 2 <£] + 2 cz sin <£ - c 2 = 0. 

Putting cos 2 <f> tan 2 <j> for sin 2 <£, we finally get, 
y 2 tan 2 -\- x 2 cos 2 <£ [tan 2 6 - tan 2 <£] + 2 cz sin <f> - c 2 = 0, 
which is the equation of the section referred to OB and OY 
in its own plane. 

Comparing this equation with the general equation of the 
second degree (Plane Geometry), we write, 

2 = cos 2 cf> tan 2 [tan 2 - tan 2 <£]. 

A = c 2 5 cos 2 <f> tan 2 [tan 2 - tan 2 <£] + tan 2 $ sin 2 0.j 

Discussion. — CI) Let c 7^ 0. 
If < 0, 2 > 0, A Tf: 0, 

.*. section is an ellipse. 

(2) If = U = O,A^ 0, 
.'. section is a parabola. 

(3) If > 8, 2 < 0, A jz 0, 
.*. section is an hyperbola. 

(4) If c = 0, A = 0, 

.*. when the secant plane passes through the vertex of the cone, the 
sections (1), (2), and (3) reduce to a point, a straight line, and two 
intersecting straight lines, respectively. 



304 ANALYTICAL GEOMETRY 

(5) If <f> = 0, secant plane is ± to axis of cone, and above equation 
becomes 

x 2 + V 2 = c 2 cot2 L a circle] 

(6) If c = oo , cone becomes a cylinder, and a section parallel to an 
element is two parallel lines; or one line when plane touches the surface. 

29. Quadrics. General equation of the second degree in 
three variables. — This may be written, 

Ax 2 + By 2 + Cz 2 + 2 Hxy + 2 Gxz + 2 ¥yz + 2 Lb 
+ 2 Jy + 2 K* + L = 0, 

and the loci which it may represent are called quadric sur- 
faces, or simply quadrics. 

Put z = A, or z = 0, and the results are conies. 

Hence, all sections || to xy plane are conies. Also, by trans- 
formation of co-ordinates, the plane xy may become one of the 
series of parallel secant planes, while the degree of the above 
equation remains unaltered. .-. all || plane sections of a quad- 
ric are similar conies. 

30. Special forms. — By transformation of co-ordinates 
the general equation may be reduced to one of the following- 
forms, viz., 

Ax 2 + By 2 + Cz 2 + L = 0, 
A* 2 + By 2 + 2 Kz = 0. 

The former represents central quadrics, the latter non- 
central quadrics. 

If the intercepts of these quadrics on the co-ordinate axes 
are a, o, and c, the former may be written in one of the three 
following forms, depending on the signs of the coefficients. 



a 2 ' b 2 ' 



x +!, + 5 = i ...... a) 



SURFACES OF REVOLUTION 305 



*+%-*±l (2) 

a 2 V & 

a*~v~& = x • (3) 

o r ,also, S + g+5-0 (4) 

^ + g-^=0 (5) 

a 2 \r & 

where the constant term L = 0. 

The second equation above, non-central quadrics, may be 
written in the two following forms : 

S+E- ^ 



but a, b, are no longer intercepts in these two equations. 

Discussions. — (1) (a) Traces on each of co-ordinate planes are 
ellipses. 

(jS) Sections || to any co-ordinate plane are similar ellipses. 
(7) Surface lies between the tangent planes 

(5) If a = b, the surface is the oblate or prolate spheroid, according 
as a > or < c. 

The surface may be generated by a variable ellipse moving || to the 
xy plane. 

It is called the ellipsoid. 

(2) (a) Trace on xy plane is an ellipse ; on yz, xz, planes, hyperbolas. 

(/3) Sections || to xy plane are similar ellipses ; those parallel to yz, xz, 
planes are hyperbolas. 

(7) Smallest elliptic section is the trace on the xy plane, semi-axes 
a and b. 

This surface is called the hyperboloid of one sheet. 

(5) If a = 6, it becomes the hyperboloid of revolution. 



306 ANALYTICAL GEOMETRY 

(3) (a) Traces on the xy and xz planes are hyperbolas. 
(/?) Sections parallel to these planes are hyperbolas. 

(7) Sections parallel to 7/2-plane are ellipses. 

(8) No part of the surface lies between the planes x = ^- a. 

(4) This equation represents the point (o, 0, 0). 

(5) This equation represents a cone, 
(a) Origin is a point on the locus. 
(/3) Trace on xy plane is a point. 

Sections parallel to xy plane are similar ellipses. 

(7) Traces on xz and yz planes are each a pair of straight lines inter- 
secting at the origin. 

(8) The surface is symmetrical with respect to each of the co-ordinate 
planes and .'. with respect to the origin as a center. Hence it is a 
central quadric. 

(6) (a) Sections parallel to xy plane are ellipses ; the xy plane is 
tangent at the origin to the surface, and the latter lies above it. 

(/S) Sections parallel to the xz and yz planes are parabolas, having the 
<r-axis as their common axis and running upward. 
This surface is known as the elliptic paraboloid. 

(7) (a) Traces on xz and yz planes are parabolas ; axes lie on z-axis, 
and they run in opposite directions. 

(|S) Sections parallel to these planes are also parabolas whose axes 
run in opposite directions. 

(8) Sections parallel to the zr/-plane are hyperbolas. Trace on this 
plane consists of two intersecting straight lines. 

Note. — Several minor details in these surfaces, such as directions of 
axes, etc., are left for the student to discuss. 
The last surface is known as the hyperbolic paraboloid. 

EXERCISES ON CHAPTER IV. 

1. Show that two spheres intersect in a circle. 

x 2 1/ 2 z 2 

2. Show that the cone -= — |= ^ = is asymptotic to the hyper- 

a 2 b 2 c 2 

boloid t ^ t l t 
a 2 b 2 c 2 

3. Show the same of the cone —. + %. = = and the hyperboloid 

a 2 b 2 c 2 

x l 4. t _ t = 1 
a? "*" b 2 c 2 



SURFACES OF REVOLUTION 307 

4. Find the plane tangent to the sphere 

(x - a) 2 + (y - h) 2 + (z - c) 2 = r 2 at (*,, y v z x ). 
Ans. (x — a) (x, — a) + (// — b) (//, — b) + (2 - c) (2! — c) = r 2 . 
Also, (s - a;,) (a - x,) + (2/ - y x ) (b - y x ) + {z - z x ) (c - z x ) = 0. 

5. Find equation of tangent plane to sphere x 2 + y 2 + z 2 = a 2 at 
(x v y v zj. Arts. xx x + yy x + z^j = a 2 . 

6. Find the equation of the tangent plane to the quadric 

Ax 2 + By 7 + Cz 2 + L = at fc, t/,, 2] ). 

Ans. Azj x 4- By, y + Cz, 2 + L = 0. 

7. Find the traces of the surface 2x 2 + Sy 2 - Gz 2 + 3 = 0. 

8. Find the projecting cylinders of the curve 

2x 2 + Sy 2 - iz 2 - 1 =0, 
Zx 2 - y 2 + z 2 + 3 = 0. 

9. Find the loci in space represented by the equations 

(1) 3 (x 2 + y 2 ) = V2. (3) 2x 2 -by 2 = 10. 

(2) Sx 2 - r oy 2 = 15. (4; (7/ - l) 2 = 12 (3 + 3). 

(5) (x 2 - 2x + 4) (x 2 + 3s 4- 1) = 0. 

EXERCISES FOR ADVANCED STUDENTS. 

1. A cone of revolution whose vertical angle = 90° is cut by a plane 
|| to one touching the slant height. Show that the latus rectum of the 
section is equal to twice its distance from the vertex. 

2. At what angle must a plane be inclined to the base of a cone in* 
order to cut a rectangular hyperbola? Determine the least vertical angle 
of the cone for which this is possible. 

3. Show that any oblique section of a cylinder of revolution is an 



4. Determine the axes of the ellipse when the radius of base of 
cylinder is given, and the inclination of the cutting plane to the axis of 
the cylinder. 

5. A right circular cylinder is cut by a plane at an angle <£ to its axis. 
Find the eccentricity of the section. Ans. e = cos <^. 



308 



ANALYTICAL GEOMETRY 




Fioi 166. 

X 2 u 2 z 2 
The Hyperboloid of one sheet, —^-\-j^ 5 = 7. 

Z 



b 2 c 2 




The Hyperboloid of two Sheets, ~ - ^ - - - 7, 
« 2 b 2 q* 



SURFACES OF REVOLUTION 



309 




Fig. 168. 



x 2 u 2 
The Elliptic Paraboloid, -= 4- ^r. = z. 
a* o* 




Fig. 169. 



X 7 u 1 z 1 

The Cone, -, + %-*- = 0. 



310 



ANALYTICAL GEOMETRY 




Fig. 170. 

The Hyperboloid and its asymptotic Cone. 




Fig. 171. 



x 1 «2 

The Hyperbolic Paraboloid, —^ — ^ = z. 



SURFACES OF REVOLUTION 



311 




Fig. 172. 



The Graph of the equations, 

y = ± b sin 2 n - ~) for two ualues of b, representing 
V the plan of the spirals of a double 
y — ±b cos 2 it - J threaded screw of pitch a. 




The Elliptic Compass. 



APPENDIX 



Note 1. — Area of a polygon in terms of the co-ordinates of its ver- 
tices (a^, 2/J, (x 2 , i/ 2 ), etc. 

Case I. Origin within the polygon. 

Area ABCDE = A OAB + A OBC + A OCD + etc. 

= i S [> 2 2/i - x x y 2 ] + [x 3 y 2 - Xtijs] + etc. J. 



(asi,Vi)j* 




Fig. 161. Fig. 162. 

Case II. Origin without the polygon. 
Area ABCDE = As OAB + OBC + OCD + OAE - ODE. 
But A ODE = - % [x h y, - x,y,\ 
Hence, in all cases, the signs of the As are positive. 
.-. Area ABCDE = ^ ][x 2 y t — x,y 2 ] + [x 3 y 2 - x 2 y 3 ] 
4- [x^y 3 -x 3 y 4 ] +etc.J 
= h \ x 2Vx - XiV* + x ^2 — x 2 y % 
x& 3 -x 3 y± -h etc. | . 
The order of the subscripts is cyclic. 

312 



APPENDIX 



313 



The student should prove the generality of this theorem by 
taking various other positions of the polygon. 
Note 2. — Eccentric angle of a point on the hyperbola. 




Fig. 163. 

Let P lie any point on the hyperbola, PN its ordinate. 

Describe the auxiliary circles, 

& + y * = a 2 (1) 

X 2^ y 2 = b 2 (2) 

From N draw NA tangent to 'circle (1). Draw OA. Then 

Z AOM = cf>, is the eccentric angle of P. 

Now, x = ON = a sec <f>. 

2/ = PN = b tan <j> 

r x 2 y 2 1 

by substitution in — — — = 1 I • 

" But BM = b tan 0. .\ BM = PN. 

The point [a sec <f>, b tan <j>] evidently satisfies the equation 
of the hyperbola 

a 2 b 2 
Hence the co-ordinates of a point on the hyperbola, as on 
the ellipse, may be given in terms of the variable parameter 
<f>, called the eccentric angle of the point. 



314 



ANALYTICAL GEOMETRY 



Note 3. — Distance between two points in terms of their polar co-ordi- 
nates, found by transformation from rectangular co-ordinates. See § 8. 

Let A be (x v y x ), B (x 2 , y 2 ). 
Then x x = p x cos d x ) x 2 = p 2 cos 2 

Vi = Pi sin *i J ' y 2 = P 2 sin 6 2 

/. AB 2 = (x 2 - x,y + (y 2 - y x f 

= [p 2 cos 6 2 — p x cos 0,] 2 + [p 2 sin 2 - ft sin X ] 2 
= ft 2 [cos 2 X + sin 2 0J + [ft 2 cos 2 6 2 + sin 2 2 ] 
— 2 ftft [cos X cos 6 2 + sin 2 sin 0,] 
= Pi 2 + P 2 ~ 2 ftft cos (0 X - d 2 ) 

Note 4. — Sign of the area of a triangle. 

See Fig. § 8, case III. Also Ex. 2, page 19. 

Area OAB = 4 OA • OB sin AOB. 
= I PJ 2 sin (0 2 - d x ) 

This result is positive or negative according as (0 2 — X ) is positive or 
negative. Hence, it is seen that if we go around the A OAB in the order 
in which 0, A, B are mentioned ; then if this order is clockwise (0 2 — X ) 
is positive ; otherwise it is negative. 

Now in Fig. 19, § 11, take A for pole, AD for initial line. Let B be 
(ft, B x ) and C (p 2 , <y. 



Then 



x x = p x COS 0j 
y x = ft sin 9 X 

= (x 2 - x i) (2/s - Vi) 
= ft cos 0j • p 2 sin 2 



p 2 cos 2 

(x 3 - x x ) {y 2 - y x ) 
p 2 cos 2 • ft sin 6 X 



= - ftft sin (d x - 2 ) 

is positive or negative according as in traversing 
the perimeter of A ABC from A to B, from B 
to C, and from C to A the order is counter- 
clockwise or clockwise. 

.-. ABC = J- £ ^ 2 ^ 2 ! 

*3 ^3 1 

the upper or lower sign prevailing according as the order is counter- 
clockwise or clockwise. If the vertices of the triangle are (o, o), (x x , y x ), 

(«2» y 2 ) • 

A = H* 2 2/i - *i 2/ 2 ] 

= ^ [ft sin ■ ft cos 2 — p x cos 0j • p 2 sin 2 ] 
= i ft ft sin (*! - 2 ) 



APPENDIX 315 

Hence, the area of the A is positive or negative according as the order 
of (o, o), (x v j/j), (x 2 , y 2 ) is clockwise or counter-clockwise. 

Note 5. — If a variable circle cut two given fixed circles at constant 
angles a and /3, then it cuts any co-axial circle at a constant angle y. 

Let the given circles be 

x 2 + y 2 + 2 G, x + 2 Fj y + C, = 0) or S, = 0, 

and x 2 + y 2 4- 2G 2 x + 2F 2 y + C 2 = 0) S 2 = 0. 

Their radii are r, and r 2 . 

Let the variable circle be 

f + j/ 2 + 2Ga; + 2F!/ + C = 0, radius K. 

Now S, + X S 2 = (radius, r) 

is any co-axial circle of the first two. 

, „G, +XG, n P,+\F 2 C. + X C 2 _ 
« ^ +;/ + 2^ r:nr a ; + 2-i n -^ !/ + ^ rT ^ = 0. 

Hence the formula for cos y gives 

2Rr cos y 1 2G V.V* 2 + 2F^ r ±¥ :2 ~ V \° 2 ~ C. 

1 -}- A 1 -f- A 1 -)- A 

.-. 2 r r 2 cos 7 (1 + /c) = 2 GG X + 2 FF X - C x - C 

+ X [2 GG 2 + 2 FF 2 - C - C] 
= 2 K r, cos a + 2 X K r 2 cos /3. 
.-. (1 + X) r, cos 7 = *i cos a + X r 2 cos /3. 
This formula shows that y is constant. 

Note 6. — A system of circles which cuts orthogonally two given circles 
has a common radical axis, i.e. is co-axial. 

Let the given circles be 

x 2 + y 2 + 2 G, x + 2 Fj i/ + C, = 
a: 2 + ?/ 2 + 2 G 2 x + 2 F 2 y + C, = 

and x 2 + i/M-2Gx + 2Ft/ + C = (1) 

one of the orthogonal circles. 

.-. 2 GG 1 + 2 FF X - C, - C = . . . . . . (2) 

2 GG 2 + 2 FF 2 - C 2 - C = (3) 

Eliminate G and F from (1), (2) and (3). 
.*. (1) may be replaced by 

x y x 2 + y 2 + C 
G i F i - Cj - C 



316 



ANALYTICAL GEOMETRY 



x y x 2 + y 2 
G x F, -C x 
G, F — Co 



+ C 



x 

G, 



y i 
f q - i 



o. 



If C varies this represents a circle of a co-axial system whose common 
radical axis is 



x y 
G, F 9 



i.e. the line joining the centers of the two given circles. 

Cor. It is easily seen that the center of the circle which cuts three 
given circles orthogonally is their radical center. 

Note 7. — The polar of a limiting point of a co-axial system is the 
same for every circle of the system. 

The polar of (k, o) with respect to the circle 
x 2 + y 2 + 2 X x + k 2 = 
is kx + \(x + k) + k 2 = 

or (x + k) (X + k) = 0, or x + k = 0, 

which is the line through the other limiting point and parallel to the 
radical axis. 

Note 8. — Joachimsthal's method of finding the equation of the tan- 
gents to a given curve drawn from an external point. 

We shall illustrate it by finding the tangents to the circle x 2 + y 2 = r 2 . 
Let (x v ?/i) be the given external point, and (;c 2 , y 2 ) any other point such 
that the line joining these is cut by the circle. Let m : n be the ratio of 
this section. Hence, the values 



mx 2 + nx 1 



V = 



my 2 + n yi 
m + n 



must satisfy 



x 2 + y 2 = r 2 



.'. (mx 2 + nXj) 2 + {my 2 + nyj 2 = r 2 (m + n) 2 . 
.'. m 2 [x 2 2 + y 2 — r 2 ] + 2 mn \x x x 2 + y x y 2 — r 2 ] +n 2 [x x 2 + y* — r 2 ] = 0. 

This quadratic determines the values of m : n corresponding to the 
two points where the line cuts the circle. It has equal roots, i.e. the line 
touches the circle if 

[x 2 + y 2 - r 2 ] [x 2 + y 2 - r 2 ] = \x x x 2 + y x y 2 - r 2 ] 2 . 



APPENDIX 



317 



This is true if (x 2 , y 2 ) is any point on either tangent from (x v y x ). 
Hence, writing x, y, for x 2 , y 2 , we obtain the equation to the pair of 
tangents from (x v y v ) viz : 

[x* + y* - &] [x* + y* - r*J = [jcr 1 + yy x - r*]\ 

This is Joachimsthal's method with some modification by the author. 
It is also applicable to the conic sections. 

Note 9. — Solution of problems in Maxima and Minima by means of 
the ellipse. 

Example 1. — Find a point P in a A ABC such that PA + PB + PC 
is a minimum. 

Suppose the sum PB + PC is given ; then P is constrained to move 
on an ellipse whose foci are B and C ; and then AP is a minimum when 
AP is a normal to the ellipse at P, and .•. makes equal angles with BP 
and CP. By symmetry, AP, BP and CP make /s of 120° with each 
other when AP + BP + CP is a minimum. P is easily found by 
Elementary Plane Geometry. The problem is impossible when any 
angle of the A ABC is greater than 120° ; but a point P can be found 
outside the A fulfilling the requirement. 

Ex. 2. — Inscribe a A DEF of minimum perimeter in the A ABC. 

Hint. — By similar reasoning show that FD and DE make equal 
angles with BC, DE and EF make equal angles with CA, and EF and 
FD make equal angles with AB. By Elementary Geometry we can 
easily show that DEF is the pedal triangle, i.e., D, E, and F are the feet 
of the _Ls from A, B, C, on the opposite sides. 

Note 10. — On Pedal curves. The locus of P, the foot of al from 
the origin on the tangent to a curve is called the pedal of the curve 




Fig. 164 a 



Fig. 164 b 



318 ANALYTICAL GEOMETRY 

with respect to 0, and is called the pole of the pedal. Thus, with 
respect to a focus, the pedal of an ellipse or hyperbola is the auxiliary 
circle, and the pedal of a parabola is the tangent at the vertex. We 
shall now find the pedal of a circle. 




Fig. 164 c 
Let C be the center, radius a, and O any point, wnere OC = b. 
Now p = OP = OL + LP = a + b cos 0. 

.-. p = a + h cos 

is the polar equation of the pedal ; the curve is the Limacon alread} r 
mentioned in this book. 

The limacon may be written 

p = J_ a + b cos 6 
corresponding to parallel tangents to the circle. Hence the chord of the 
liniacon through O is of constant length 2 a. 

Discussion of the equation. 

(1) If b > a, O is outside the circle and the pedal is looped, having a 
double point O [Fig. 6]. 

(2) If b = a, is on the circle, and 

p = a (1 + cos 6) 
which is the cardioid [Fig. c], 

(3) If b < a, is inside the circle and the pedal consists of a single 
oval curve [Fig. a]. 

This oval has points of inflexion if b > \ a. 

The limacon is sometimes called the conchoid of the circle because it 
is described by producing the radius vector of the circle [/> = b cos 6] a 
constant distance a. 



APPENDIX 



319 



Note 11. — The slope or gradient of the tangent to a curve at the point 
(x, y) on the curve. 

If (x, y) be any point on a curve and {x+Ax, y+Ay) another point 
on the curve near the point (x, y) then the slope or gradient of the tan- 
gent drawn to the curve at the point (x, y) is equal to limiting value of 

Ay 
the fraction -p when the second point is made to approach the first 

and, ultimately, to coincide with it. 

n 



(x*Ax,y*Ay) Q 




fay) % 



u 



Fig. 174. 



Thus let PQ be a small arc of a curve and let the co-ordinates of P 
be (x, y) and those of Q be {x-\-Ax, y-\-Ay). Draw the ordinates QR 
and PS, and draw PN perpendicular to QR. Then we have 



tan <f> 



QN = Ay 
PN~Ax' 



the slope of the chord PQ. Now suppose the point Q to approach 
indefinitely near to P. As Q moves into coincidence with P the chord 
PQ becomes the tangent PT and the slope of the tangent becomes 



320 ANALYTICAL GEOMETRY 



A?/ 
the imiting value of — . Hence in finding the equation of a tangent 

Ay 
to a curve we shall have to determine the limit of — when the two 

Ax 

points on the curve approach each other and ultimately coincide. 

Suppose we are to find the equation of the tangent at the particular 

point (xi, yi) on a curve. Let (xi-\-Ax h yi-j-Ayi) be another point on 

the curve near the first point. Then the equation of the chord joining 

the two points is 

u-y X-Zi 



yi+Ayx-yi. Xi+Azj— xi' 
or 

y-yi jc-xx ^ 
Ayi Axi ' 
i.e., 

y-^ = ATS x ~ Xl) ' 

This equation shows that the fraction ■— is the slope or m of the chord 

(see § 24) and therefore its limiting value is the slope of the tangent 
to the curve at the point (x u y x ). This result does not hold when the 
axes of co-ordinates are oblique. 

Note 12. — To find the equation of the tangent to the circle x 2 -\-y 2 =r 2 at 
the point (xi, y{). The method employed in § 59 is known as the 
secant method. We shall now exemplify the method of increments; 
explained in the preceding note. Let (zi+Axi, y^+Ayi) be another 
point on the circle near the given point. Then, both these points being 
on the curve, their co-ordinates must satisfy the given equation. There- 
fore 

Xl 2 + 2/l 2 =r 2 ........ (1) 

and 

(*! -1^)2 + (?/i +A?/ 1 ) 2 = r? (2) 

Expanding, and subtracting (1) from (2), we obtain 

2^*1 + (AxO 2 +2y x A yi + (A^) 2 = 0. 

.*. proceeding to the limit, when the points coincide, and neglecting 
(Azi) 2 and (A?/i) 2 , we have 



APPENDIX 321 



2x 1 Ax x +2 h -Ay i =Q 

Ayi_ _xi 

Ax x y x > 
Xt 

i.e.. — is the slope or m of the tangent at Xi, y\. 

Also, since the tangent passes through (xi, y{) its equation is 

Xi. , 

[another way of writing y — yi = m(x — X\)\ 
or 

= ^2 4-1/, 2 

/. xxi+yy, =r 2 

is the required equation of the tangent, which agrees with the result 
found in § 59. 

Third Method. Use of the Differential Calculus. 

Differentiating the equation 

X 2+y2 =r 2 

with respect to x, we get 

2*+2j,|=0, 

. dy _ _s 
" dx y 

X\ 

-is the slope of the tangent at (xi, yi). 

• • y-yi = -—(x-^x l ) 

or 

or 

xx l +yy l = r 2 

is the required equation of the tangent, as previously found. 



322 ANALYTICAL GEOMETRY 

Note 13. — To find the equation of the tangent to the circle 

a; 2 +2/ 2 +2Gx+2F?/+C =0. 

The equation found in § 60 will now be derived by the method of incre- 
ments. Take the two points (x\, yi) and {xi-\-Ax xi yi+Ayi) on the circle, 
near each other. Then, expressing the fact that each lies on the 
curve, we have 

Xi*+y l *+2Gx 1 +2Fy l +C = (1) 

(x 1 +Ax 1 )*+(y 1 +Ayi)*+2G(x 1 +Ax 1 )+2F(y 1 +Ay 1 )+C = . . (2) 
whence, by subtraction 

2x x Ax* + (Azi) 2 +2y 1 Ay 1 + {Ay, ) 2 +2GAx : +2FA2/ x =0. 
Taking limits, 

.'. Ax 1 (x 1 +G)+Ay l (y 1 +F) =0, 

and the limiting value of 

Ay i _zH-G 
Axi~ 2/i+F 

^Ppr is the slope of the tangent at (xi, yi). 

Substituting this for m in the equation 
y— y x =m(x—xi) 



we get 



or 






xx } +yyi +Gx+Fy = x l *+y i *+Gxi+Fy 1 
or 

xXi+yyi+Gx+Fy = -Gxx-Fyx — C 

[since (xi, yi) lies on the circle] 

or 

xx 1 -Vyyi+G(x+x 1 )+F(y+y l )+C=0 

which is the required equation of the tangent. 



APPENDIX 323 

Third Method by Means of the Differential Calculus. 

Differentiating the equation with respect to x, we have 

2, + 2,| + 2G + 2F|=0 

dy _ x-\-G 
" dx~~y + F 

.'. the slope of the tangent at (x h y x ) is — -^ 
Whence 

.'. ^i+yz/i+G(a;+x 1 )+F(2/+2/i)+C=0. 
Note 14. — To find the equation of the tangent to the circle. 

x 2 -\-y 2 -\-2xy cos 4> = r 2 

at the point (xi, t/0, where the co-ordinate axes are oblique and include 
an angle </>. (See § 57.) 

Let (.ri+Axi, 2/i+A?/i) be another point on the circle near (xi, ij\). 
The equation of the chord joining the two points is, [by note 11, above] 

y-y l _x-x l 
Ayi AX! Kl} 

Also, since both points lie on the circle, 

Xi*+y i 2 +2x 1 y 1 coa <f>=r* (2) 

and 

(x 1 +Ax 1 y + (y l +Ay l ) 2 +2(x l +Ax l )(i Jl +Ay l ) cos 4>=r\ . (3) 

by subtraction, 

2x l Ax 1 + (Ax l ) 2 +2y l Ay l + (Ay l )*+2{y l Ax i +x l Ay l + Ax 1 Aij l ) c s <£=0. 

In the limit, we may neglect the small quantities (AxO 2 , (AyyO 2 and 
AxrA?/i, thus obtaining 

XiAXi+?yiA?yi + (2/iA;ri+£iA?/i) cos = 

.*. Ai/i^i-r-XiCOS (/>) = —Axifa+yx cos <j>). ... (4) 



324 ANALYTICAL GEOMETRY 

Multiplying equations (1) and (4) together we get, eliminating Axi, 
and Ayi, 

(y—yi)(yi+xi cos <j>) = —(x — xjfa+yi cos 4>) 

which is the equation of the tangent, and may be written 

xx 1 -\-yy 1 -\-{x x y-\ r yix) cos </>=:r 1 2 +?/ 1 2 +2:i-i2/i cos 

.*. xx! +yyi +(xiy+yix) cos </> = r 2 . 

The equation x 2 -\-y 2 -\-2xy cos <f>=r 2 , above, may be obtained from 
the more general equation in § 57 by taking the center at the origin 
and making h = 0, k = 0. 

Note 15. — To find the equation of the tangent at the point (xi, y{) on 
the parabola y 2 =^ax. 

In § 102 we found this equation by the secant method. 
Let (xi, yi) and (xi+Axi, yi-\-Ayi) be neighboring points on the 
curve. Then 

?/i 2 = 4axi (1) 

(y 1 +Ay l ) 2 = 4:a(x 1 +Ax 1 ) (2) 

or 

l/ 1 2 +2ij 1 Ay l + (Ay 1 ) 2 = 4:ax 1 +4:aAx 1 (3) 

From (1) and (3) by subtraction, 

2?/iA?/i+A7/i 2 = 4aA.Ti 

Proceeding to the limit, where the points coincide, 
ijxAyi = 2aAxi 
Aj/i_ 2a 

Am ~ yi 
.'. — is the slope of the tangent. 

yi 

Also, since the tangent passes through (x h yi) its equation is 

y-y l = m(x-xi) 

2a. . 

i.e. y-y 1 = — {x-xi) 

Vi 
or 

yyi —yi 2 = 2ax — 2ax\ 



APPENDIX 325 

since the point (xi, 7/1) is on the curve, t/i 2 =4<2:ei, and the pre- 
ceding equation becomes 

7/2/1 — 4azi = 2ax 2axi 

:. yy 1 =2a(x+xi) 
is the required equation. 

Third Method. The Differential Calculus. 

Differentiating the equation y 2 = £ax with respect to x, we obtain 

J dx 

dy _ 2a 
" dx y 

which is the slope of the tangent at any point {x, y) on the curve. 

the slope of the tangent at (x\, t/i) is — . 

Substituting this value for m in the equation 

y-yi=m{x-xx) 

and reducing as before, we have again 

T/?/i= 2a(x+xi). 

Similarly the tangent to the parabola x 2 = 4a7/, at the point (xi, 7/0 can 
be found, viz : 

xx y =2a(ij+yi). 

In § 116 it was shown that the equation of the parabola when referred 
to any diameter and the tangent at its extremity as axes, takes a similar 

form, viz: 

7/ 2 = 4a'x. 

By a method analogous to that employed above it can be shown that 
the tangent to this curve at (xi, t/i) is 

yy 1 =2a'(x+xi). 

Note 16. — To find the equation of the tangent to the ellipse, 



ai+t^ 1 W 



at the point (x h yj. [See § 127.] 



326 ANALYTICAL GEOMETRY 

Let (zi-f-Azi, i^-j-Ai/i) be another point on the curve, close by the 
given point. 

Also let the required equation of the tangent be 

y—yi=m(x—xi) (2) 

A?/i 
where m is the limiting value of ~r~. 

Axi 

Now, since the two points are on the curve,, we have by equation (1), 

£+£-» ^ 

a 2 "^ 6 2 w 



By subtraction, 



Taking limits 



2x 1 Ax 1 + (A.t 1 ) 2 2y 1 Ay 1 + (A?/ 1 ) 2 

a 2 "+" 6 2 U# 



xxbxx yiAyi 

,2 h2 W > 



a 2 ' 6 2 

. Ayi = b*xi 

Axi a V 

which is the required slope of the tangent. Substituting this value fo* 
m in equation (2), we have 

. yyi yi 2 _ xxi x Y 2 

" b 2 b 2 a 2_h a 2 

or 

zzi Wi = »£ , yi* 

a 2 + 6 2 a 2_t ~ 6 2 ' 

in which the right-hand member is, by equation (3), equal to 1. 

_<7 I 7 o -*- 



6- 



is the equation of the tangent. 

Third Method. The Differential Calculus. Differentiating the 
equation of the ellipse with respect to x, we obtain 



APPENDIX 327 

2z,2j/^/ 
a 2 ~^b 2 dx 

dy _ bhc 
dx a 2 y' 

which is the slope of the tangent at any point (x, y) on the curve. For 
the particular point (x h y{) we have 

b*x x 

m= — . 

a 2 yi 

Proceeding, as before, with equation (2) we obtain the equation of 
the tangent. 

Note 17. — To find the equation of the tangent to the hyperbola 

x*_y* 

a 2 b 2 

at the point (x\, yi). 

The work of the preceding note holds in this case if — b 2 is written 
in place of b 2 . The required equation is found by both the method 
of increments and the differential calculus to be 

xxi W\ = * 

a 2 b 2 

Note 18. — To find the equation of the tangent to the hyperbola 2xy = c 2 
at the point (x h yi). The equation found in § 165 is sometimes written 
in this form 

Let y — y\=m(x—xi) (1) 

be the required equation of the tangent and let (xi+Axi, ?/i+A?/i) be 
another point near the first on the curve. Then 

2x lVl =c 2 (2) 

and 

2(x 1 +Ax 1 )(y l +Ai Jl )=c 2 (3) 

By subtraction 



XiAyi -\-yiAxi + AxiAyi = 0. 

Taking limits, 

/. XiAy 1 +yiAx i =0 



A?/i y\ 

Axi X\ 



328 ANALYTICAL GEOMETRY 

Putting this value in (1), we have 

.'. xiy+yix =2xiy t 
xnj-\-yiX =c 2 

is the required equation. 

This equation may be written in another form. Thus, dividing by 
Xiz/i, we obtain 



or, finally 



y ■ x = c i = c* =2 

2/i a;i Xi?/i |c 2 ' 



*-+^ = 2. 
£i 2/i 



Note 19. — To find the equation of the tangent at the point (x h y{) to 
the general conic 

Az 2 +2H.n/+B7/ 2 +2G:r+2F?/+C=0 (1) 

We shall now derive equation (1) of § 190 by the method of incre- 
ments and the differential calculus. 

Let (:ri+A:ri, ?/i+A?/i) be a second point on the conic near to the 
given point. 

Then, the two points being on the conic, their co-ordinates satisfy 
the given equation, 

/. Ax 1 2 +2B.x 1 y l +By 1 *+2Gx l +2Fy l +C = 
and 

A(x 1 +Ax 1 )*+2H.(x l +Ax l )(y 1 +Ay l )+B(y 1 +Ay l y+2G(x l +Ax 1 ) 

+2F( 2/l +A 2 / 1 )+c=0. 

Subtracting and neglecting terms containing (A:ri) 2 , (At/i) 2 , Axi-Aj/i, 
we have 

2Ax l Ax 1 +2Ti(y l Ax 1 +x 1 Ay 1 ) +2By 1 Ay 1 +2GAz 1 +2FAj/ 1 =0. 

Taking limits 

• A2/!_ Asi+H yi+G 
" Axi Kxt+Byi+F m 

/. the equation of the tangent is 



APPENDIX 329 

Axi+Hyt+G, 
y - yi= -Jix 1+ By 1+ F {x - Xl) ' 

which may be written 

Axx 1 +B.(x l y+y l x)+Byy 1 +Gx+Fy = Ax i *+2JIx 1 y 1 +By 1 z+Gx 1 +Fy l . 
But the right-hand member equals — Gx x — Fyi — C from equation (1). 

/. Axx 1 +K(xy 1 +x 1 y)+Byy 1 +G(x+x 1 )-\-F(y+y l )+C=0 

is the required equation of the tangent. 

To obtain this result by the differential calculus, differentiate the 
given equation with respect to x, 

.'. (2Hz+2B^+2F)^4-2A:c4-2H?/+2G = 

dy_ A.r+H?/+G 
" dx~ Ux+By+F 

which is the slope of the tangent at any point. For the point (xi, y x ) 
the slope becomes 

Aji+Hyi+G _ 

H^+Byi + F ™" 

Proceeding as before, we readily obtain the preceding result. 

Note 20. — On the area of a triangle in polar co-ordinates. In § 12 
we found the area of the triangle ABC to be 

= \[nn sin (0i - 2 ) +r 2 r 3 sin (0 2 - 3 ) +r 3 n sin (0 3 - 0i)] 



inr 2 r 3 



= -2-rir 2 r 3 



sin(0i-0 2 ) . sin (0>-0 3 ) sin 



+ 



n 

sin 0i 
sin 2 
sin 0a 



+ 



(01-03) 1 

n J 



COS 0\ 
COS 02 
COS 3 



This may be easily verified by expanding the determinant and com- 
bining terms. Compare this with the result found in § 40. Hence, 
the condition for three colli near points in polar co-ordinates is the van- 
ishing of this determinant, i.e., 



330 



ANALYTICAL GEOMETRY 



n 


sin X 


cos 6i 


1 
r 2 


sin 02 


COS 02 


1 


sin 3 


cos 3 



Note 21. — 0?i Simson's Line. The feet of the perpendiculars 
drawn from any point O on the circumference of a circle to the sides of 
an inscribed triangle ABC are collinear. 

Employing polar co-ordinates, let O be the pole, and the diameter 
through O the initial line. Also let the vectorial angles of the vertices 
A, B, C, be a, /3, y. Then the equation of the circle is 



r = 2a cos 



(1) 



Then the co-ordinates of B and C are (2a cos /3, /3) and (2i cos 7, 7). 
To find the polar equation of BC, take the form derived in § 41, namaly 

p=r cos (d-4>) (2) 

and substitute therein the co-ordinates of B and C. We get two 
equations involving p and <p, viz : 



and 

Solving these, 
and 



p = 2acos/3cos (/3-</>) (3) 

p = 2a cos 7 cos (7 — 0) (4) 

/. <£=/3+7 (5) 

p = 2a cos /3 cos 7 (6) 

substituting these values in (2), we get as the equation of BC 

2a cos /? cos y=r cos (0-/3 — 7) (7) 

Similarly we find the equations of CA and AB, viz: 

2a cos 7 cos a = r cos (0 — 7— a) (8) 

2a cos a cos /3=r cos (0— a— /?) (9) 

The equation of the perpendicular from O on BC is 
0=r sin (0-/3-7). 



and 



APPENDIX 



331 



The co-ordinates of the foot of this perpendicular are therefore 

(2a cos cos y, fi+y). 

Similarly the feet of th > other two perpendiculars are found to be 
(2a cos 7 cos a, y-\-a) and (2a cos a cos 0, a+0). If these -diree 
points are collinear then the following determinant must vanish, as 
proved in Note 20 above. 

Thus the necessary condition is 



1 



2a COS a cos /3 cos 7 



2a cos a cos 

1 





2a cos 



1 


COS 


7 


2a cos 7 


cos a 


cos 7 








COS a. 








cos 









sin (a+0) cos (a+0) 
sin (0+7) cos (0+7) 
sin (7+a) COS (7+ a) 



= 



sin (a+0) cos (a+0) 1 
sin (0+7) cos (0+7) =0 
sin (7+a) cos (7+a) I 



Let a +/3 = .r, 
minant becomes 



+7=2/, y+a = z, «+0+7 = m. Then the deter- 

cos (u — x) sin x cos 2; 

cos (u — y) sin ?/ cos y 

cos (u— 2) sin 2 cos 2 

cos (u— x) sin (y— 2)+cos (n — y) sin (2— x)+cos (w— 2) sin (x — y) = 
cos 7 sin (0— a)+cos a sin (7— 0)+cos sin (a — 7)=0 

which proves the proposition. 

Again, the condition of collinearity found above can also be obtained 
from the polar equation of the straight line, § 40. Thus if the points 
Cn, 0i ), (r 2 , 2 ), and (r 3 , 3 ) are collinear, the co-ordinates (r 3 , 3 ) must 
satisfy the determinant found there, 



cos 



sin 



cos 0i sin 0i 
cos 2 sin 2 



= 



332 



ANALYTICAL GEOMETRY 



sin di 



n 
1_ 
r 2 
1 






cos 2 


sin 2 


COS 3 


sin 3 



Again, the polar equation of Simson's Line is found to be 

2a cos a cos j3 cos y=r cos (0—a — p — y). 

This is evidently an example where the methods of Pure Geometry 
are more useful. 



Note 22. — To find the condition that the straight line 
lx-\-my = n 



(1) 

may touch the conic 

A.x 2 +2H.T?/+B?y 2 = C (2) 

By finding the value of y from (1) and substituting it in (2), we get an 
equation of the second degree in x. which determines the abscissas of 
the points of intersection of the straight line and the curve, viz: 



m m l 



(3 



The straight line will be a tangent when the two roots of (3) are equal. 
The condition for equal roots is that the expression 

(Am 2 - 2HZm +Bl 2 )x 2 +2(Hmn - Bln)x 4-Bn 2 - Cm 2 
be a perfect square. » 

.\ (Hmn-BZn) 2 = (Bn 2 -Cm 2 )(Ara 2 -2HZm+BZ 2 ) 
/. ACm 2 +BC/ 2 -(AB-H 2 )n 2 -2CHZm=0 



0, 



is the required condition. 

Note 23. — To determine the circle orthogonal to the three circles 

(x-a^ + iy-b^^n* (1) 



A 


H 


I 


H 


B 


in 


CI 


Cm 


n 2 



APPENDIX 



333 



(x-a 2 ) 2 + (y-b 2 ) 2 = r 2 2 (2) 

(x-a3) 2 +(7,-6 3 ) 2 =r3 2 (3) 

[See § 83.] 

Let the required equation be 

(x-a) 2 +(y-b) 2 =r 2 (4) 

Then, since it cuts each given circle at right angles, the square of the 
distance between the centers is equal to the sum of the squares of the radii. 

.-. (a-a l ) 2 +(b-b i ) 2 =r 2 +r l 2 

i.e., a 2 +b 2 -r 2 -2aa i -2bb l +a l 2 +b i 2 -r i 2 = (5; 

Similarly, we find 



and 



a 2 +6 2 







(6) 
(7) 



a 2 +fr 2 -r 2 -2aa 3 -26& 3 +a, 2 +&3 2 -r 3 2 = 0. . . . 

Writing equation (4) in expanded form we have 

a*+b 2 -r 2 -2ax-2by+x 2 +y 2 = (8) 

Now, regarding equations (5), (6), (7), (8) as simultaneous in 
a 2 -\-b 2 — r 2 , a, b, and eliminating these quantities, we have 







x 2 +y 2 , 
a 1 2 +6i 2 -r 1 2 , 
a 2 2 +b 2 2 -r 2 2 , 
aj+bj-n 2 , 
as the equation of the required circle. 

Note 24. — To find the equation of the straight line joining the point of 
intersection of the lines 



X, 


y, 


l 


(h, 


K 


l 


a 2> 


h 


l 


a 3 , 


■ K 


l 



and that of the lines 



aix-\~biy+ci=0 

a 2 x-\-boy-\-c 2 =Q 

liX-\-m\y-\-n\=0 
l 2 x+ni2y-\-no=0 



[See page 82, ex. 11. 
The equation 

faidiX+biy+ci) +h(aeX+b2y+Ci) =0 



(1) 
(2) 

(3) 
(4) 

(5) 



334 ANALYTICAL GEOMETRY 

represents a straight line through the intersection of (1) and (2) because 
it is of the first degree, and is satisfied by the co-ordinates of their 
common point. Now, in order that it may pass also through the 
intersection of (3) and (4), the co-ordinates of this point also must 
satisfy the equation. On substituting the co-ordinates of this point 
in (5) we get 

ki[ai(min 2 — m 2 nx) +bx(nxl 2 —n 2 lx) +Cx{lxm2 — Unix)] + 

~k 2 [a 2 {mxn 2 — rrhnx) + t 2 (niZ 2 — n 2 h ) +c 2 {lxm 2 — l 2 mx)] = 0. 

This equation gives a value for the ratio k h /k 2 . Substituting this value 
in equation (5), we have the equation of the required line, viz: 

[a 2 (mxn 2 —m 2 ni)+b 2 (nxl 2 -n 2 l 1 )+c 2 (lxm 2 -l 2 mx)](axX+bxy+Cx) 

= [ax {mxn 2 — m 2 nx ) + bx (nik — n 2 lx) + Cx (lxm 2 — l 2 rrix ) ] {a^x -\- b 2 y + c 2 ) 



a 2 b 2 <h 
lx mx fix 
h m 2 n 2 



(axx+bxy + ci) 



ax 


bx 


Cx 


lx 


mx 


nx 


h 


m 2 


n 2 



(a 2 x+biy+c 2 ) 



Note 25. — Several misprints occur in Note 5, above. The last three 
lines there should read as follows : 

.*. 2Rrcos 7 (l+X)=2GG 1 +2FF 1 -C 1 -C+X(2GG J +2FF 2 -C 2 -C) 

= 2Rr x cos a+2XRr 2 cos /3 

(1+X)r cos y—tx, cos a+\r 2 cos /3. 

This formula shows that y is constant. 

Note £6. — To find the condition that the straight line 

Ix+my + 1=0 (1) 

may be a tangent to the general conic 

A.x 2 +2H.n/+£?/ 2 +2G.r+2Fy+C=0 (2) 

Suppose the point of contact of (1) and (2) to be the point (xx, yi). Then 
the equation (1) must be identical with the equation of the tangent 
to the conic (2) at the point (xx, yx)', that is, with the equation, 

(Axx+TIyx+G)x+(Rxx+Byx+F)y + (Gxx+Fyx+C) =0 . (3) 

[See § 190, equation (1).] 



APPENDIX 



335 



Hence the coefficients in equations (1) and (3) must be proportional, 

/. i = X(Aafc+Hi&+G) (4) 

m = \(Bx 1 +By l +¥) (5) 

l=X(Ga;i+Fyi+C) (6) 

Now, multiplying (4) by X\ and (5) by yi and adding them to (6), 
we have 



lx 1 +my 1 +l=\(Ax 1 *+2E.x 1 y 1 +ByS+2Gx 1 +2Fy l +C). 



(7) 



But since the point (x h y{) is on the curve (2), the right member of (7) 
vanishes. 

.'. Ixx +my x +1=0, (8) 

or 

X(Zx 1 +m 2/l + l)=0 (9) 

Now, equations (4), (5), (6), and (9) are a system of four simultaneous 
equations of the first degree in the variables X.ri, \y h and X. The elim- 
ination of these three quantities gives the required condition, viz: 



A 


H 


G 


I 


H 


B 


F 


m 


G 


F 


C 


i 


I 


m 


i 


o 



Remark. — The development of this determinant gives: 
A/ 2 +2HZra+B™ 2 +2GZ+2Fm+C =0. 

Note 27. — The intersections of a parabola and a circle. 
[See page 155, ex. 3 and page 269, ex. 98.] 

The points of intersection of the circle 

x 2 +?/ 2 +2Gr+2F?/+C=0 . . , 
and the parabola 

?/ 2 =4a.r , 

are found by eliminating x from (1) and (2). 

^ 4 ; 2 +?/ 2 + 2 ^ 2 +2F 2/ +c=o, 



(1) 

(2) 



or 



16a 2 ' " ' 4a 
?/ 4 +8a(2a+G)?/ 2 +32a 2 F?/ + 16a 2 C =0. 
This is a biauadratic equation in y and has four roots; but by the 



336 ANALYTICAL GEOMETRY 

theory of equations the sum of these four values is zero since the 
coefficient of y 3 is zero. 

Note 28. — Solution of example 9, page 70. 

Let AB=a, AD = b, AG = m, AF = n. Then the equation of FH 
which joins F(n, 0) and H(a, m) is 

V = m 
x—n a—n 
or 

ay—ny = mx — mn (1) 

Similarly, the equation of GE, joining E(n, 6) and G(p, m) is 

y — b_ m—b 
x—n — n 
or 

bn—ny=mx — bx—mn-{-bn 
or 

bx — ny = mx—mn (2) 

By subtracting (2) from (1) we get a line which must pass through their 
intersection, for all values of m and n. [See § 49, in which let k= — 1.] 

ay—bx=0 

is the required locus and evidently is the equation of the diagonal AC. 

Note 29. — In the result of example 31, page 85, the denominator of 
the right-hand member should read 

AZ 2 +2HZm+Bn 2 

and the query is not proper as n does enter into the answer. 

Note 30. The following discussion, by analysis, of the relation 
between two circles in a plane may prove interesting to students, as 
an instance where analytical methods are useful. 

Let R and r be the radii of the two circles, and d the distance between 
their centers. Take the center of the first circle as the origin of rectang- 
lar co-ordinates, and the line through the centers as the x-axis. 

Then the equations of the two circles are 

x*+y*=R\ (1) 

(x-dy+y* = r* (2> 



APPENDIX 337 



Solving (1) and (2) for x and y, we get 
R 2 -r 2 +d 2 



2d 



(3) 



7/ = ±^V4R 2 <2 2 -(R 2 -r 2 +d 2 ) 2 ; 

= ±^rV(2Rd+R*-r 2 +d 2 ) (2Rd-R*+r 2 -d 2 ) 
1 



= ±2^V{ (R+d) 2 -r 2 } {r 2 - (R-d) 2 j 



or, 



= ±^V(R+d+r)(R+d-r)(r+R-d)(r-R+d) 



y=d=^V(R+r+d)(R-r+d)(R+r-d)(r-R+d). (4) 



An examination of (3) and (4) shows that: 

1. The abscissa x is always real. 

2. If the continued product under the radical sign in (4) is positive, 
then the two points of intersection of the circles are real. These two 
points have the same abscissa, and two ordinates equal, but of opposite 
signs. Hence, when two circles intersect, their common chord is bisected 
perpendicularly by the line through their centers. 

3. Since the first factor under the radical sign is always positive, the 
two values of y will be real if (a) the three other factors are positive, 
or (6) one of them is positive and the remaining two negative. But 
condition (6) cannot exist; for if one of the three said factors be negative, 
the remaining two must necessarily be positive. Thus, supposing the 
second factor, for example, to be negative, then we have 

R-r+d<0; 

.'. R+d<r; 
hence 

d<r, (5) 

and 

R<r (6) 

Hence (5) shows that the third factor must be positive and (6) shows that 
the fourth factor is positive. Similarly, if either the third or fourth 
factor be supposed negative the remaining two can be shown to be posi- 
tive. Furthermore, the three stated factors cannot all be negative, 



338 ANALYTICAL GEOMETRY 

together. Hence, the only supposition tenable (for both values of y 
to be real) is (a) above; i.e., all four factors must be -positive. Therefore: 

4. Two circles intersect in two real -points when the sum of each two of 
the three quantities R, r, d, is greater than the third. 

5. If any one of the three said factors be zero, we get y = 0, and the 
two circumferences have only one common point, which lies on the 
x-axis; i.e., the point of contact of two tangent circles lies on the line 
through the centers. 

G. The first factor, obviously, cannot vanish. Hence, if the two cir- 
cles are to have only one point in common, i.e., are to touch each other 
then the second, third or fourth factor must vanish. 

If the second or fourth factor vanishes, 

R-r=±d, 
or 

d = ±(R-r) (7) 

If the third factor vanishes, - 

d = R+r (8) 

Therefore : 

7. Two circles will touch each other when the distance between their 
c orders is equal to the sum or difference of their radii. 

8. If any one of the three factors mentioned be negative, the re- 
maining two are positive, by 3. In that case the values of y will be 
imaginary; i.e., the circumferences will not meet. Hence: 

9. The circumferences will not meet if 

R-r+d<0, i.e., r-R>d, (9) 

or 

R+r-d<0, i.e., R+r<d, (10) 

or 

r-R+d<0, i.e., R-r>d. ..... (11) 

Therefore, by (9) and (11), the circumferences will not meet when the 
difference of the radii is greater than the distance between the centers, 
i.e., in that case one circle lies wholly within the other. Also, by (10), 
the circles will not meet when the sum of the radii is less than the dis- 
tance between the centers; in which case, the circles are externally; 
apart from each other. 



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